-0.000 000 000 742 147 676 646 713 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 713 7(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
-0.000 000 000 742 147 676 646 713 7(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:

|-0.000 000 000 742 147 676 646 713 7| = 0.000 000 000 742 147 676 646 713 7


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: 0.000 000 000 742 147 676 646 713 7.

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) 0.000 000 000 742 147 676 646 713 7 × 2 = 0 + 0.000 000 001 484 295 353 293 427 4;
  • 2) 0.000 000 001 484 295 353 293 427 4 × 2 = 0 + 0.000 000 002 968 590 706 586 854 8;
  • 3) 0.000 000 002 968 590 706 586 854 8 × 2 = 0 + 0.000 000 005 937 181 413 173 709 6;
  • 4) 0.000 000 005 937 181 413 173 709 6 × 2 = 0 + 0.000 000 011 874 362 826 347 419 2;
  • 5) 0.000 000 011 874 362 826 347 419 2 × 2 = 0 + 0.000 000 023 748 725 652 694 838 4;
  • 6) 0.000 000 023 748 725 652 694 838 4 × 2 = 0 + 0.000 000 047 497 451 305 389 676 8;
  • 7) 0.000 000 047 497 451 305 389 676 8 × 2 = 0 + 0.000 000 094 994 902 610 779 353 6;
  • 8) 0.000 000 094 994 902 610 779 353 6 × 2 = 0 + 0.000 000 189 989 805 221 558 707 2;
  • 9) 0.000 000 189 989 805 221 558 707 2 × 2 = 0 + 0.000 000 379 979 610 443 117 414 4;
  • 10) 0.000 000 379 979 610 443 117 414 4 × 2 = 0 + 0.000 000 759 959 220 886 234 828 8;
  • 11) 0.000 000 759 959 220 886 234 828 8 × 2 = 0 + 0.000 001 519 918 441 772 469 657 6;
  • 12) 0.000 001 519 918 441 772 469 657 6 × 2 = 0 + 0.000 003 039 836 883 544 939 315 2;
  • 13) 0.000 003 039 836 883 544 939 315 2 × 2 = 0 + 0.000 006 079 673 767 089 878 630 4;
  • 14) 0.000 006 079 673 767 089 878 630 4 × 2 = 0 + 0.000 012 159 347 534 179 757 260 8;
  • 15) 0.000 012 159 347 534 179 757 260 8 × 2 = 0 + 0.000 024 318 695 068 359 514 521 6;
  • 16) 0.000 024 318 695 068 359 514 521 6 × 2 = 0 + 0.000 048 637 390 136 719 029 043 2;
  • 17) 0.000 048 637 390 136 719 029 043 2 × 2 = 0 + 0.000 097 274 780 273 438 058 086 4;
  • 18) 0.000 097 274 780 273 438 058 086 4 × 2 = 0 + 0.000 194 549 560 546 876 116 172 8;
  • 19) 0.000 194 549 560 546 876 116 172 8 × 2 = 0 + 0.000 389 099 121 093 752 232 345 6;
  • 20) 0.000 389 099 121 093 752 232 345 6 × 2 = 0 + 0.000 778 198 242 187 504 464 691 2;
  • 21) 0.000 778 198 242 187 504 464 691 2 × 2 = 0 + 0.001 556 396 484 375 008 929 382 4;
  • 22) 0.001 556 396 484 375 008 929 382 4 × 2 = 0 + 0.003 112 792 968 750 017 858 764 8;
  • 23) 0.003 112 792 968 750 017 858 764 8 × 2 = 0 + 0.006 225 585 937 500 035 717 529 6;
  • 24) 0.006 225 585 937 500 035 717 529 6 × 2 = 0 + 0.012 451 171 875 000 071 435 059 2;
  • 25) 0.012 451 171 875 000 071 435 059 2 × 2 = 0 + 0.024 902 343 750 000 142 870 118 4;
  • 26) 0.024 902 343 750 000 142 870 118 4 × 2 = 0 + 0.049 804 687 500 000 285 740 236 8;
  • 27) 0.049 804 687 500 000 285 740 236 8 × 2 = 0 + 0.099 609 375 000 000 571 480 473 6;
  • 28) 0.099 609 375 000 000 571 480 473 6 × 2 = 0 + 0.199 218 750 000 001 142 960 947 2;
  • 29) 0.199 218 750 000 001 142 960 947 2 × 2 = 0 + 0.398 437 500 000 002 285 921 894 4;
  • 30) 0.398 437 500 000 002 285 921 894 4 × 2 = 0 + 0.796 875 000 000 004 571 843 788 8;
  • 31) 0.796 875 000 000 004 571 843 788 8 × 2 = 1 + 0.593 750 000 000 009 143 687 577 6;
  • 32) 0.593 750 000 000 009 143 687 577 6 × 2 = 1 + 0.187 500 000 000 018 287 375 155 2;
  • 33) 0.187 500 000 000 018 287 375 155 2 × 2 = 0 + 0.375 000 000 000 036 574 750 310 4;
  • 34) 0.375 000 000 000 036 574 750 310 4 × 2 = 0 + 0.750 000 000 000 073 149 500 620 8;
  • 35) 0.750 000 000 000 073 149 500 620 8 × 2 = 1 + 0.500 000 000 000 146 299 001 241 6;
  • 36) 0.500 000 000 000 146 299 001 241 6 × 2 = 1 + 0.000 000 000 000 292 598 002 483 2;
  • 37) 0.000 000 000 000 292 598 002 483 2 × 2 = 0 + 0.000 000 000 000 585 196 004 966 4;
  • 38) 0.000 000 000 000 585 196 004 966 4 × 2 = 0 + 0.000 000 000 001 170 392 009 932 8;
  • 39) 0.000 000 000 001 170 392 009 932 8 × 2 = 0 + 0.000 000 000 002 340 784 019 865 6;
  • 40) 0.000 000 000 002 340 784 019 865 6 × 2 = 0 + 0.000 000 000 004 681 568 039 731 2;
  • 41) 0.000 000 000 004 681 568 039 731 2 × 2 = 0 + 0.000 000 000 009 363 136 079 462 4;
  • 42) 0.000 000 000 009 363 136 079 462 4 × 2 = 0 + 0.000 000 000 018 726 272 158 924 8;
  • 43) 0.000 000 000 018 726 272 158 924 8 × 2 = 0 + 0.000 000 000 037 452 544 317 849 6;
  • 44) 0.000 000 000 037 452 544 317 849 6 × 2 = 0 + 0.000 000 000 074 905 088 635 699 2;
  • 45) 0.000 000 000 074 905 088 635 699 2 × 2 = 0 + 0.000 000 000 149 810 177 271 398 4;
  • 46) 0.000 000 000 149 810 177 271 398 4 × 2 = 0 + 0.000 000 000 299 620 354 542 796 8;
  • 47) 0.000 000 000 299 620 354 542 796 8 × 2 = 0 + 0.000 000 000 599 240 709 085 593 6;
  • 48) 0.000 000 000 599 240 709 085 593 6 × 2 = 0 + 0.000 000 001 198 481 418 171 187 2;
  • 49) 0.000 000 001 198 481 418 171 187 2 × 2 = 0 + 0.000 000 002 396 962 836 342 374 4;
  • 50) 0.000 000 002 396 962 836 342 374 4 × 2 = 0 + 0.000 000 004 793 925 672 684 748 8;
  • 51) 0.000 000 004 793 925 672 684 748 8 × 2 = 0 + 0.000 000 009 587 851 345 369 497 6;
  • 52) 0.000 000 009 587 851 345 369 497 6 × 2 = 0 + 0.000 000 019 175 702 690 738 995 2;
  • 53) 0.000 000 019 175 702 690 738 995 2 × 2 = 0 + 0.000 000 038 351 405 381 477 990 4;
  • 54) 0.000 000 038 351 405 381 477 990 4 × 2 = 0 + 0.000 000 076 702 810 762 955 980 8;

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:


0.000 000 000 742 147 676 646 713 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 713 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 646 713 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 713 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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