-0.000 000 000 000 151 582 441 260 469 917 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 151 582 441 260 469 917 5(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
-0.000 000 000 000 151 582 441 260 469 917 5(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:

|-0.000 000 000 000 151 582 441 260 469 917 5| = 0.000 000 000 000 151 582 441 260 469 917 5


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: 0.000 000 000 000 151 582 441 260 469 917 5.

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) 0.000 000 000 000 151 582 441 260 469 917 5 × 2 = 0 + 0.000 000 000 000 303 164 882 520 939 835;
  • 2) 0.000 000 000 000 303 164 882 520 939 835 × 2 = 0 + 0.000 000 000 000 606 329 765 041 879 67;
  • 3) 0.000 000 000 000 606 329 765 041 879 67 × 2 = 0 + 0.000 000 000 001 212 659 530 083 759 34;
  • 4) 0.000 000 000 001 212 659 530 083 759 34 × 2 = 0 + 0.000 000 000 002 425 319 060 167 518 68;
  • 5) 0.000 000 000 002 425 319 060 167 518 68 × 2 = 0 + 0.000 000 000 004 850 638 120 335 037 36;
  • 6) 0.000 000 000 004 850 638 120 335 037 36 × 2 = 0 + 0.000 000 000 009 701 276 240 670 074 72;
  • 7) 0.000 000 000 009 701 276 240 670 074 72 × 2 = 0 + 0.000 000 000 019 402 552 481 340 149 44;
  • 8) 0.000 000 000 019 402 552 481 340 149 44 × 2 = 0 + 0.000 000 000 038 805 104 962 680 298 88;
  • 9) 0.000 000 000 038 805 104 962 680 298 88 × 2 = 0 + 0.000 000 000 077 610 209 925 360 597 76;
  • 10) 0.000 000 000 077 610 209 925 360 597 76 × 2 = 0 + 0.000 000 000 155 220 419 850 721 195 52;
  • 11) 0.000 000 000 155 220 419 850 721 195 52 × 2 = 0 + 0.000 000 000 310 440 839 701 442 391 04;
  • 12) 0.000 000 000 310 440 839 701 442 391 04 × 2 = 0 + 0.000 000 000 620 881 679 402 884 782 08;
  • 13) 0.000 000 000 620 881 679 402 884 782 08 × 2 = 0 + 0.000 000 001 241 763 358 805 769 564 16;
  • 14) 0.000 000 001 241 763 358 805 769 564 16 × 2 = 0 + 0.000 000 002 483 526 717 611 539 128 32;
  • 15) 0.000 000 002 483 526 717 611 539 128 32 × 2 = 0 + 0.000 000 004 967 053 435 223 078 256 64;
  • 16) 0.000 000 004 967 053 435 223 078 256 64 × 2 = 0 + 0.000 000 009 934 106 870 446 156 513 28;
  • 17) 0.000 000 009 934 106 870 446 156 513 28 × 2 = 0 + 0.000 000 019 868 213 740 892 313 026 56;
  • 18) 0.000 000 019 868 213 740 892 313 026 56 × 2 = 0 + 0.000 000 039 736 427 481 784 626 053 12;
  • 19) 0.000 000 039 736 427 481 784 626 053 12 × 2 = 0 + 0.000 000 079 472 854 963 569 252 106 24;
  • 20) 0.000 000 079 472 854 963 569 252 106 24 × 2 = 0 + 0.000 000 158 945 709 927 138 504 212 48;
  • 21) 0.000 000 158 945 709 927 138 504 212 48 × 2 = 0 + 0.000 000 317 891 419 854 277 008 424 96;
  • 22) 0.000 000 317 891 419 854 277 008 424 96 × 2 = 0 + 0.000 000 635 782 839 708 554 016 849 92;
  • 23) 0.000 000 635 782 839 708 554 016 849 92 × 2 = 0 + 0.000 001 271 565 679 417 108 033 699 84;
  • 24) 0.000 001 271 565 679 417 108 033 699 84 × 2 = 0 + 0.000 002 543 131 358 834 216 067 399 68;
  • 25) 0.000 002 543 131 358 834 216 067 399 68 × 2 = 0 + 0.000 005 086 262 717 668 432 134 799 36;
  • 26) 0.000 005 086 262 717 668 432 134 799 36 × 2 = 0 + 0.000 010 172 525 435 336 864 269 598 72;
  • 27) 0.000 010 172 525 435 336 864 269 598 72 × 2 = 0 + 0.000 020 345 050 870 673 728 539 197 44;
  • 28) 0.000 020 345 050 870 673 728 539 197 44 × 2 = 0 + 0.000 040 690 101 741 347 457 078 394 88;
  • 29) 0.000 040 690 101 741 347 457 078 394 88 × 2 = 0 + 0.000 081 380 203 482 694 914 156 789 76;
  • 30) 0.000 081 380 203 482 694 914 156 789 76 × 2 = 0 + 0.000 162 760 406 965 389 828 313 579 52;
  • 31) 0.000 162 760 406 965 389 828 313 579 52 × 2 = 0 + 0.000 325 520 813 930 779 656 627 159 04;
  • 32) 0.000 325 520 813 930 779 656 627 159 04 × 2 = 0 + 0.000 651 041 627 861 559 313 254 318 08;
  • 33) 0.000 651 041 627 861 559 313 254 318 08 × 2 = 0 + 0.001 302 083 255 723 118 626 508 636 16;
  • 34) 0.001 302 083 255 723 118 626 508 636 16 × 2 = 0 + 0.002 604 166 511 446 237 253 017 272 32;
  • 35) 0.002 604 166 511 446 237 253 017 272 32 × 2 = 0 + 0.005 208 333 022 892 474 506 034 544 64;
  • 36) 0.005 208 333 022 892 474 506 034 544 64 × 2 = 0 + 0.010 416 666 045 784 949 012 069 089 28;
  • 37) 0.010 416 666 045 784 949 012 069 089 28 × 2 = 0 + 0.020 833 332 091 569 898 024 138 178 56;
  • 38) 0.020 833 332 091 569 898 024 138 178 56 × 2 = 0 + 0.041 666 664 183 139 796 048 276 357 12;
  • 39) 0.041 666 664 183 139 796 048 276 357 12 × 2 = 0 + 0.083 333 328 366 279 592 096 552 714 24;
  • 40) 0.083 333 328 366 279 592 096 552 714 24 × 2 = 0 + 0.166 666 656 732 559 184 193 105 428 48;
  • 41) 0.166 666 656 732 559 184 193 105 428 48 × 2 = 0 + 0.333 333 313 465 118 368 386 210 856 96;
  • 42) 0.333 333 313 465 118 368 386 210 856 96 × 2 = 0 + 0.666 666 626 930 236 736 772 421 713 92;
  • 43) 0.666 666 626 930 236 736 772 421 713 92 × 2 = 1 + 0.333 333 253 860 473 473 544 843 427 84;
  • 44) 0.333 333 253 860 473 473 544 843 427 84 × 2 = 0 + 0.666 666 507 720 946 947 089 686 855 68;
  • 45) 0.666 666 507 720 946 947 089 686 855 68 × 2 = 1 + 0.333 333 015 441 893 894 179 373 711 36;
  • 46) 0.333 333 015 441 893 894 179 373 711 36 × 2 = 0 + 0.666 666 030 883 787 788 358 747 422 72;
  • 47) 0.666 666 030 883 787 788 358 747 422 72 × 2 = 1 + 0.333 332 061 767 575 576 717 494 845 44;
  • 48) 0.333 332 061 767 575 576 717 494 845 44 × 2 = 0 + 0.666 664 123 535 151 153 434 989 690 88;
  • 49) 0.666 664 123 535 151 153 434 989 690 88 × 2 = 1 + 0.333 328 247 070 302 306 869 979 381 76;
  • 50) 0.333 328 247 070 302 306 869 979 381 76 × 2 = 0 + 0.666 656 494 140 604 613 739 958 763 52;
  • 51) 0.666 656 494 140 604 613 739 958 763 52 × 2 = 1 + 0.333 312 988 281 209 227 479 917 527 04;
  • 52) 0.333 312 988 281 209 227 479 917 527 04 × 2 = 0 + 0.666 625 976 562 418 454 959 835 054 08;
  • 53) 0.666 625 976 562 418 454 959 835 054 08 × 2 = 1 + 0.333 251 953 124 836 909 919 670 108 16;
  • 54) 0.333 251 953 124 836 909 919 670 108 16 × 2 = 0 + 0.666 503 906 249 673 819 839 340 216 32;
  • 55) 0.666 503 906 249 673 819 839 340 216 32 × 2 = 1 + 0.333 007 812 499 347 639 678 680 432 64;
  • 56) 0.333 007 812 499 347 639 678 680 432 64 × 2 = 0 + 0.666 015 624 998 695 279 357 360 865 28;
  • 57) 0.666 015 624 998 695 279 357 360 865 28 × 2 = 1 + 0.332 031 249 997 390 558 714 721 730 56;
  • 58) 0.332 031 249 997 390 558 714 721 730 56 × 2 = 0 + 0.664 062 499 994 781 117 429 443 461 12;
  • 59) 0.664 062 499 994 781 117 429 443 461 12 × 2 = 1 + 0.328 124 999 989 562 234 858 886 922 24;
  • 60) 0.328 124 999 989 562 234 858 886 922 24 × 2 = 0 + 0.656 249 999 979 124 469 717 773 844 48;
  • 61) 0.656 249 999 979 124 469 717 773 844 48 × 2 = 1 + 0.312 499 999 958 248 939 435 547 688 96;
  • 62) 0.312 499 999 958 248 939 435 547 688 96 × 2 = 0 + 0.624 999 999 916 497 878 871 095 377 92;
  • 63) 0.624 999 999 916 497 878 871 095 377 92 × 2 = 1 + 0.249 999 999 832 995 757 742 190 755 84;
  • 64) 0.249 999 999 832 995 757 742 190 755 84 × 2 = 0 + 0.499 999 999 665 991 515 484 381 511 68;
  • 65) 0.499 999 999 665 991 515 484 381 511 68 × 2 = 0 + 0.999 999 999 331 983 030 968 763 023 36;
  • 66) 0.999 999 999 331 983 030 968 763 023 36 × 2 = 1 + 0.999 999 998 663 966 061 937 526 046 72;

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:


0.000 000 000 000 151 582 441 260 469 917 5(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:

0.000 000 000 000 151 582 441 260 469 917 5(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:


0.000 000 000 000 151 582 441 260 469 917 5(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 -0.000 000 000 000 151 582 441 260 469 917 5 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