-0.000 000 000 000 151 582 441 260 469 923 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 151 582 441 260 469 923 3(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 000 151 582 441 260 469 923 3(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 000 151 582 441 260 469 923 3| = 0.000 000 000 000 151 582 441 260 469 923 3


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 151 582 441 260 469 923 3.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 000 151 582 441 260 469 923 3 × 2 = 0 + 0.000 000 000 000 303 164 882 520 939 846 6;
  • 2) 0.000 000 000 000 303 164 882 520 939 846 6 × 2 = 0 + 0.000 000 000 000 606 329 765 041 879 693 2;
  • 3) 0.000 000 000 000 606 329 765 041 879 693 2 × 2 = 0 + 0.000 000 000 001 212 659 530 083 759 386 4;
  • 4) 0.000 000 000 001 212 659 530 083 759 386 4 × 2 = 0 + 0.000 000 000 002 425 319 060 167 518 772 8;
  • 5) 0.000 000 000 002 425 319 060 167 518 772 8 × 2 = 0 + 0.000 000 000 004 850 638 120 335 037 545 6;
  • 6) 0.000 000 000 004 850 638 120 335 037 545 6 × 2 = 0 + 0.000 000 000 009 701 276 240 670 075 091 2;
  • 7) 0.000 000 000 009 701 276 240 670 075 091 2 × 2 = 0 + 0.000 000 000 019 402 552 481 340 150 182 4;
  • 8) 0.000 000 000 019 402 552 481 340 150 182 4 × 2 = 0 + 0.000 000 000 038 805 104 962 680 300 364 8;
  • 9) 0.000 000 000 038 805 104 962 680 300 364 8 × 2 = 0 + 0.000 000 000 077 610 209 925 360 600 729 6;
  • 10) 0.000 000 000 077 610 209 925 360 600 729 6 × 2 = 0 + 0.000 000 000 155 220 419 850 721 201 459 2;
  • 11) 0.000 000 000 155 220 419 850 721 201 459 2 × 2 = 0 + 0.000 000 000 310 440 839 701 442 402 918 4;
  • 12) 0.000 000 000 310 440 839 701 442 402 918 4 × 2 = 0 + 0.000 000 000 620 881 679 402 884 805 836 8;
  • 13) 0.000 000 000 620 881 679 402 884 805 836 8 × 2 = 0 + 0.000 000 001 241 763 358 805 769 611 673 6;
  • 14) 0.000 000 001 241 763 358 805 769 611 673 6 × 2 = 0 + 0.000 000 002 483 526 717 611 539 223 347 2;
  • 15) 0.000 000 002 483 526 717 611 539 223 347 2 × 2 = 0 + 0.000 000 004 967 053 435 223 078 446 694 4;
  • 16) 0.000 000 004 967 053 435 223 078 446 694 4 × 2 = 0 + 0.000 000 009 934 106 870 446 156 893 388 8;
  • 17) 0.000 000 009 934 106 870 446 156 893 388 8 × 2 = 0 + 0.000 000 019 868 213 740 892 313 786 777 6;
  • 18) 0.000 000 019 868 213 740 892 313 786 777 6 × 2 = 0 + 0.000 000 039 736 427 481 784 627 573 555 2;
  • 19) 0.000 000 039 736 427 481 784 627 573 555 2 × 2 = 0 + 0.000 000 079 472 854 963 569 255 147 110 4;
  • 20) 0.000 000 079 472 854 963 569 255 147 110 4 × 2 = 0 + 0.000 000 158 945 709 927 138 510 294 220 8;
  • 21) 0.000 000 158 945 709 927 138 510 294 220 8 × 2 = 0 + 0.000 000 317 891 419 854 277 020 588 441 6;
  • 22) 0.000 000 317 891 419 854 277 020 588 441 6 × 2 = 0 + 0.000 000 635 782 839 708 554 041 176 883 2;
  • 23) 0.000 000 635 782 839 708 554 041 176 883 2 × 2 = 0 + 0.000 001 271 565 679 417 108 082 353 766 4;
  • 24) 0.000 001 271 565 679 417 108 082 353 766 4 × 2 = 0 + 0.000 002 543 131 358 834 216 164 707 532 8;
  • 25) 0.000 002 543 131 358 834 216 164 707 532 8 × 2 = 0 + 0.000 005 086 262 717 668 432 329 415 065 6;
  • 26) 0.000 005 086 262 717 668 432 329 415 065 6 × 2 = 0 + 0.000 010 172 525 435 336 864 658 830 131 2;
  • 27) 0.000 010 172 525 435 336 864 658 830 131 2 × 2 = 0 + 0.000 020 345 050 870 673 729 317 660 262 4;
  • 28) 0.000 020 345 050 870 673 729 317 660 262 4 × 2 = 0 + 0.000 040 690 101 741 347 458 635 320 524 8;
  • 29) 0.000 040 690 101 741 347 458 635 320 524 8 × 2 = 0 + 0.000 081 380 203 482 694 917 270 641 049 6;
  • 30) 0.000 081 380 203 482 694 917 270 641 049 6 × 2 = 0 + 0.000 162 760 406 965 389 834 541 282 099 2;
  • 31) 0.000 162 760 406 965 389 834 541 282 099 2 × 2 = 0 + 0.000 325 520 813 930 779 669 082 564 198 4;
  • 32) 0.000 325 520 813 930 779 669 082 564 198 4 × 2 = 0 + 0.000 651 041 627 861 559 338 165 128 396 8;
  • 33) 0.000 651 041 627 861 559 338 165 128 396 8 × 2 = 0 + 0.001 302 083 255 723 118 676 330 256 793 6;
  • 34) 0.001 302 083 255 723 118 676 330 256 793 6 × 2 = 0 + 0.002 604 166 511 446 237 352 660 513 587 2;
  • 35) 0.002 604 166 511 446 237 352 660 513 587 2 × 2 = 0 + 0.005 208 333 022 892 474 705 321 027 174 4;
  • 36) 0.005 208 333 022 892 474 705 321 027 174 4 × 2 = 0 + 0.010 416 666 045 784 949 410 642 054 348 8;
  • 37) 0.010 416 666 045 784 949 410 642 054 348 8 × 2 = 0 + 0.020 833 332 091 569 898 821 284 108 697 6;
  • 38) 0.020 833 332 091 569 898 821 284 108 697 6 × 2 = 0 + 0.041 666 664 183 139 797 642 568 217 395 2;
  • 39) 0.041 666 664 183 139 797 642 568 217 395 2 × 2 = 0 + 0.083 333 328 366 279 595 285 136 434 790 4;
  • 40) 0.083 333 328 366 279 595 285 136 434 790 4 × 2 = 0 + 0.166 666 656 732 559 190 570 272 869 580 8;
  • 41) 0.166 666 656 732 559 190 570 272 869 580 8 × 2 = 0 + 0.333 333 313 465 118 381 140 545 739 161 6;
  • 42) 0.333 333 313 465 118 381 140 545 739 161 6 × 2 = 0 + 0.666 666 626 930 236 762 281 091 478 323 2;
  • 43) 0.666 666 626 930 236 762 281 091 478 323 2 × 2 = 1 + 0.333 333 253 860 473 524 562 182 956 646 4;
  • 44) 0.333 333 253 860 473 524 562 182 956 646 4 × 2 = 0 + 0.666 666 507 720 947 049 124 365 913 292 8;
  • 45) 0.666 666 507 720 947 049 124 365 913 292 8 × 2 = 1 + 0.333 333 015 441 894 098 248 731 826 585 6;
  • 46) 0.333 333 015 441 894 098 248 731 826 585 6 × 2 = 0 + 0.666 666 030 883 788 196 497 463 653 171 2;
  • 47) 0.666 666 030 883 788 196 497 463 653 171 2 × 2 = 1 + 0.333 332 061 767 576 392 994 927 306 342 4;
  • 48) 0.333 332 061 767 576 392 994 927 306 342 4 × 2 = 0 + 0.666 664 123 535 152 785 989 854 612 684 8;
  • 49) 0.666 664 123 535 152 785 989 854 612 684 8 × 2 = 1 + 0.333 328 247 070 305 571 979 709 225 369 6;
  • 50) 0.333 328 247 070 305 571 979 709 225 369 6 × 2 = 0 + 0.666 656 494 140 611 143 959 418 450 739 2;
  • 51) 0.666 656 494 140 611 143 959 418 450 739 2 × 2 = 1 + 0.333 312 988 281 222 287 918 836 901 478 4;
  • 52) 0.333 312 988 281 222 287 918 836 901 478 4 × 2 = 0 + 0.666 625 976 562 444 575 837 673 802 956 8;
  • 53) 0.666 625 976 562 444 575 837 673 802 956 8 × 2 = 1 + 0.333 251 953 124 889 151 675 347 605 913 6;
  • 54) 0.333 251 953 124 889 151 675 347 605 913 6 × 2 = 0 + 0.666 503 906 249 778 303 350 695 211 827 2;
  • 55) 0.666 503 906 249 778 303 350 695 211 827 2 × 2 = 1 + 0.333 007 812 499 556 606 701 390 423 654 4;
  • 56) 0.333 007 812 499 556 606 701 390 423 654 4 × 2 = 0 + 0.666 015 624 999 113 213 402 780 847 308 8;
  • 57) 0.666 015 624 999 113 213 402 780 847 308 8 × 2 = 1 + 0.332 031 249 998 226 426 805 561 694 617 6;
  • 58) 0.332 031 249 998 226 426 805 561 694 617 6 × 2 = 0 + 0.664 062 499 996 452 853 611 123 389 235 2;
  • 59) 0.664 062 499 996 452 853 611 123 389 235 2 × 2 = 1 + 0.328 124 999 992 905 707 222 246 778 470 4;
  • 60) 0.328 124 999 992 905 707 222 246 778 470 4 × 2 = 0 + 0.656 249 999 985 811 414 444 493 556 940 8;
  • 61) 0.656 249 999 985 811 414 444 493 556 940 8 × 2 = 1 + 0.312 499 999 971 622 828 888 987 113 881 6;
  • 62) 0.312 499 999 971 622 828 888 987 113 881 6 × 2 = 0 + 0.624 999 999 943 245 657 777 974 227 763 2;
  • 63) 0.624 999 999 943 245 657 777 974 227 763 2 × 2 = 1 + 0.249 999 999 886 491 315 555 948 455 526 4;
  • 64) 0.249 999 999 886 491 315 555 948 455 526 4 × 2 = 0 + 0.499 999 999 772 982 631 111 896 911 052 8;
  • 65) 0.499 999 999 772 982 631 111 896 911 052 8 × 2 = 0 + 0.999 999 999 545 965 262 223 793 822 105 6;
  • 66) 0.999 999 999 545 965 262 223 793 822 105 6 × 2 = 1 + 0.999 999 999 091 930 524 447 587 644 211 2;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 000 151 582 441 260 469 923 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1010 1010 1010 1010 01(2)

6. Positive number before normalization:

0.000 000 000 000 151 582 441 260 469 923 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1010 1010 1010 1010 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 151 582 441 260 469 923 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1010 1010 1010 1010 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1010 1010 1010 1010 01(2) × 20 =

1.0101 0101 0101 0101 0101 001(2) × 2-43


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -43

Mantissa (not normalized):
1.0101 0101 0101 0101 0101 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-43 + 2(8-1) - 1 =

(-43 + 127)(10) =

84(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 84 ÷ 2 = 42 + 0;
  • 42 ÷ 2 = 21 + 0;
  • 21 ÷ 2 = 10 + 1;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

84(10) =

0101 0100(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 010 1010 1010 1010 1010 1001 =

010 1010 1010 1010 1010 1001


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0101 0100

Mantissa (23 bits) =
010 1010 1010 1010 1010 1001


Decimal number -0.000 000 000 000 151 582 441 260 469 923 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 0100 - 010 1010 1010 1010 1010 1001

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111