-0.000 000 000 742 147 676 646 584 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 584(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
-0.000 000 000 742 147 676 646 584(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:

|-0.000 000 000 742 147 676 646 584| = 0.000 000 000 742 147 676 646 584


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: 0.000 000 000 742 147 676 646 584.

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) 0.000 000 000 742 147 676 646 584 × 2 = 0 + 0.000 000 001 484 295 353 293 168;
  • 2) 0.000 000 001 484 295 353 293 168 × 2 = 0 + 0.000 000 002 968 590 706 586 336;
  • 3) 0.000 000 002 968 590 706 586 336 × 2 = 0 + 0.000 000 005 937 181 413 172 672;
  • 4) 0.000 000 005 937 181 413 172 672 × 2 = 0 + 0.000 000 011 874 362 826 345 344;
  • 5) 0.000 000 011 874 362 826 345 344 × 2 = 0 + 0.000 000 023 748 725 652 690 688;
  • 6) 0.000 000 023 748 725 652 690 688 × 2 = 0 + 0.000 000 047 497 451 305 381 376;
  • 7) 0.000 000 047 497 451 305 381 376 × 2 = 0 + 0.000 000 094 994 902 610 762 752;
  • 8) 0.000 000 094 994 902 610 762 752 × 2 = 0 + 0.000 000 189 989 805 221 525 504;
  • 9) 0.000 000 189 989 805 221 525 504 × 2 = 0 + 0.000 000 379 979 610 443 051 008;
  • 10) 0.000 000 379 979 610 443 051 008 × 2 = 0 + 0.000 000 759 959 220 886 102 016;
  • 11) 0.000 000 759 959 220 886 102 016 × 2 = 0 + 0.000 001 519 918 441 772 204 032;
  • 12) 0.000 001 519 918 441 772 204 032 × 2 = 0 + 0.000 003 039 836 883 544 408 064;
  • 13) 0.000 003 039 836 883 544 408 064 × 2 = 0 + 0.000 006 079 673 767 088 816 128;
  • 14) 0.000 006 079 673 767 088 816 128 × 2 = 0 + 0.000 012 159 347 534 177 632 256;
  • 15) 0.000 012 159 347 534 177 632 256 × 2 = 0 + 0.000 024 318 695 068 355 264 512;
  • 16) 0.000 024 318 695 068 355 264 512 × 2 = 0 + 0.000 048 637 390 136 710 529 024;
  • 17) 0.000 048 637 390 136 710 529 024 × 2 = 0 + 0.000 097 274 780 273 421 058 048;
  • 18) 0.000 097 274 780 273 421 058 048 × 2 = 0 + 0.000 194 549 560 546 842 116 096;
  • 19) 0.000 194 549 560 546 842 116 096 × 2 = 0 + 0.000 389 099 121 093 684 232 192;
  • 20) 0.000 389 099 121 093 684 232 192 × 2 = 0 + 0.000 778 198 242 187 368 464 384;
  • 21) 0.000 778 198 242 187 368 464 384 × 2 = 0 + 0.001 556 396 484 374 736 928 768;
  • 22) 0.001 556 396 484 374 736 928 768 × 2 = 0 + 0.003 112 792 968 749 473 857 536;
  • 23) 0.003 112 792 968 749 473 857 536 × 2 = 0 + 0.006 225 585 937 498 947 715 072;
  • 24) 0.006 225 585 937 498 947 715 072 × 2 = 0 + 0.012 451 171 874 997 895 430 144;
  • 25) 0.012 451 171 874 997 895 430 144 × 2 = 0 + 0.024 902 343 749 995 790 860 288;
  • 26) 0.024 902 343 749 995 790 860 288 × 2 = 0 + 0.049 804 687 499 991 581 720 576;
  • 27) 0.049 804 687 499 991 581 720 576 × 2 = 0 + 0.099 609 374 999 983 163 441 152;
  • 28) 0.099 609 374 999 983 163 441 152 × 2 = 0 + 0.199 218 749 999 966 326 882 304;
  • 29) 0.199 218 749 999 966 326 882 304 × 2 = 0 + 0.398 437 499 999 932 653 764 608;
  • 30) 0.398 437 499 999 932 653 764 608 × 2 = 0 + 0.796 874 999 999 865 307 529 216;
  • 31) 0.796 874 999 999 865 307 529 216 × 2 = 1 + 0.593 749 999 999 730 615 058 432;
  • 32) 0.593 749 999 999 730 615 058 432 × 2 = 1 + 0.187 499 999 999 461 230 116 864;
  • 33) 0.187 499 999 999 461 230 116 864 × 2 = 0 + 0.374 999 999 998 922 460 233 728;
  • 34) 0.374 999 999 998 922 460 233 728 × 2 = 0 + 0.749 999 999 997 844 920 467 456;
  • 35) 0.749 999 999 997 844 920 467 456 × 2 = 1 + 0.499 999 999 995 689 840 934 912;
  • 36) 0.499 999 999 995 689 840 934 912 × 2 = 0 + 0.999 999 999 991 379 681 869 824;
  • 37) 0.999 999 999 991 379 681 869 824 × 2 = 1 + 0.999 999 999 982 759 363 739 648;
  • 38) 0.999 999 999 982 759 363 739 648 × 2 = 1 + 0.999 999 999 965 518 727 479 296;
  • 39) 0.999 999 999 965 518 727 479 296 × 2 = 1 + 0.999 999 999 931 037 454 958 592;
  • 40) 0.999 999 999 931 037 454 958 592 × 2 = 1 + 0.999 999 999 862 074 909 917 184;
  • 41) 0.999 999 999 862 074 909 917 184 × 2 = 1 + 0.999 999 999 724 149 819 834 368;
  • 42) 0.999 999 999 724 149 819 834 368 × 2 = 1 + 0.999 999 999 448 299 639 668 736;
  • 43) 0.999 999 999 448 299 639 668 736 × 2 = 1 + 0.999 999 998 896 599 279 337 472;
  • 44) 0.999 999 998 896 599 279 337 472 × 2 = 1 + 0.999 999 997 793 198 558 674 944;
  • 45) 0.999 999 997 793 198 558 674 944 × 2 = 1 + 0.999 999 995 586 397 117 349 888;
  • 46) 0.999 999 995 586 397 117 349 888 × 2 = 1 + 0.999 999 991 172 794 234 699 776;
  • 47) 0.999 999 991 172 794 234 699 776 × 2 = 1 + 0.999 999 982 345 588 469 399 552;
  • 48) 0.999 999 982 345 588 469 399 552 × 2 = 1 + 0.999 999 964 691 176 938 799 104;
  • 49) 0.999 999 964 691 176 938 799 104 × 2 = 1 + 0.999 999 929 382 353 877 598 208;
  • 50) 0.999 999 929 382 353 877 598 208 × 2 = 1 + 0.999 999 858 764 707 755 196 416;
  • 51) 0.999 999 858 764 707 755 196 416 × 2 = 1 + 0.999 999 717 529 415 510 392 832;
  • 52) 0.999 999 717 529 415 510 392 832 × 2 = 1 + 0.999 999 435 058 831 020 785 664;
  • 53) 0.999 999 435 058 831 020 785 664 × 2 = 1 + 0.999 998 870 117 662 041 571 328;
  • 54) 0.999 998 870 117 662 041 571 328 × 2 = 1 + 0.999 997 740 235 324 083 142 656;

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:


0.000 000 000 742 147 676 646 584(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 584(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 646 584(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

96(10) =

0110 0000(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. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 646 584 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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