-0.000 000 000 742 147 677 85 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 85(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
-0.000 000 000 742 147 677 85(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:

|-0.000 000 000 742 147 677 85| = 0.000 000 000 742 147 677 85


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: 0.000 000 000 742 147 677 85.

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) 0.000 000 000 742 147 677 85 × 2 = 0 + 0.000 000 001 484 295 355 7;
  • 2) 0.000 000 001 484 295 355 7 × 2 = 0 + 0.000 000 002 968 590 711 4;
  • 3) 0.000 000 002 968 590 711 4 × 2 = 0 + 0.000 000 005 937 181 422 8;
  • 4) 0.000 000 005 937 181 422 8 × 2 = 0 + 0.000 000 011 874 362 845 6;
  • 5) 0.000 000 011 874 362 845 6 × 2 = 0 + 0.000 000 023 748 725 691 2;
  • 6) 0.000 000 023 748 725 691 2 × 2 = 0 + 0.000 000 047 497 451 382 4;
  • 7) 0.000 000 047 497 451 382 4 × 2 = 0 + 0.000 000 094 994 902 764 8;
  • 8) 0.000 000 094 994 902 764 8 × 2 = 0 + 0.000 000 189 989 805 529 6;
  • 9) 0.000 000 189 989 805 529 6 × 2 = 0 + 0.000 000 379 979 611 059 2;
  • 10) 0.000 000 379 979 611 059 2 × 2 = 0 + 0.000 000 759 959 222 118 4;
  • 11) 0.000 000 759 959 222 118 4 × 2 = 0 + 0.000 001 519 918 444 236 8;
  • 12) 0.000 001 519 918 444 236 8 × 2 = 0 + 0.000 003 039 836 888 473 6;
  • 13) 0.000 003 039 836 888 473 6 × 2 = 0 + 0.000 006 079 673 776 947 2;
  • 14) 0.000 006 079 673 776 947 2 × 2 = 0 + 0.000 012 159 347 553 894 4;
  • 15) 0.000 012 159 347 553 894 4 × 2 = 0 + 0.000 024 318 695 107 788 8;
  • 16) 0.000 024 318 695 107 788 8 × 2 = 0 + 0.000 048 637 390 215 577 6;
  • 17) 0.000 048 637 390 215 577 6 × 2 = 0 + 0.000 097 274 780 431 155 2;
  • 18) 0.000 097 274 780 431 155 2 × 2 = 0 + 0.000 194 549 560 862 310 4;
  • 19) 0.000 194 549 560 862 310 4 × 2 = 0 + 0.000 389 099 121 724 620 8;
  • 20) 0.000 389 099 121 724 620 8 × 2 = 0 + 0.000 778 198 243 449 241 6;
  • 21) 0.000 778 198 243 449 241 6 × 2 = 0 + 0.001 556 396 486 898 483 2;
  • 22) 0.001 556 396 486 898 483 2 × 2 = 0 + 0.003 112 792 973 796 966 4;
  • 23) 0.003 112 792 973 796 966 4 × 2 = 0 + 0.006 225 585 947 593 932 8;
  • 24) 0.006 225 585 947 593 932 8 × 2 = 0 + 0.012 451 171 895 187 865 6;
  • 25) 0.012 451 171 895 187 865 6 × 2 = 0 + 0.024 902 343 790 375 731 2;
  • 26) 0.024 902 343 790 375 731 2 × 2 = 0 + 0.049 804 687 580 751 462 4;
  • 27) 0.049 804 687 580 751 462 4 × 2 = 0 + 0.099 609 375 161 502 924 8;
  • 28) 0.099 609 375 161 502 924 8 × 2 = 0 + 0.199 218 750 323 005 849 6;
  • 29) 0.199 218 750 323 005 849 6 × 2 = 0 + 0.398 437 500 646 011 699 2;
  • 30) 0.398 437 500 646 011 699 2 × 2 = 0 + 0.796 875 001 292 023 398 4;
  • 31) 0.796 875 001 292 023 398 4 × 2 = 1 + 0.593 750 002 584 046 796 8;
  • 32) 0.593 750 002 584 046 796 8 × 2 = 1 + 0.187 500 005 168 093 593 6;
  • 33) 0.187 500 005 168 093 593 6 × 2 = 0 + 0.375 000 010 336 187 187 2;
  • 34) 0.375 000 010 336 187 187 2 × 2 = 0 + 0.750 000 020 672 374 374 4;
  • 35) 0.750 000 020 672 374 374 4 × 2 = 1 + 0.500 000 041 344 748 748 8;
  • 36) 0.500 000 041 344 748 748 8 × 2 = 1 + 0.000 000 082 689 497 497 6;
  • 37) 0.000 000 082 689 497 497 6 × 2 = 0 + 0.000 000 165 378 994 995 2;
  • 38) 0.000 000 165 378 994 995 2 × 2 = 0 + 0.000 000 330 757 989 990 4;
  • 39) 0.000 000 330 757 989 990 4 × 2 = 0 + 0.000 000 661 515 979 980 8;
  • 40) 0.000 000 661 515 979 980 8 × 2 = 0 + 0.000 001 323 031 959 961 6;
  • 41) 0.000 001 323 031 959 961 6 × 2 = 0 + 0.000 002 646 063 919 923 2;
  • 42) 0.000 002 646 063 919 923 2 × 2 = 0 + 0.000 005 292 127 839 846 4;
  • 43) 0.000 005 292 127 839 846 4 × 2 = 0 + 0.000 010 584 255 679 692 8;
  • 44) 0.000 010 584 255 679 692 8 × 2 = 0 + 0.000 021 168 511 359 385 6;
  • 45) 0.000 021 168 511 359 385 6 × 2 = 0 + 0.000 042 337 022 718 771 2;
  • 46) 0.000 042 337 022 718 771 2 × 2 = 0 + 0.000 084 674 045 437 542 4;
  • 47) 0.000 084 674 045 437 542 4 × 2 = 0 + 0.000 169 348 090 875 084 8;
  • 48) 0.000 169 348 090 875 084 8 × 2 = 0 + 0.000 338 696 181 750 169 6;
  • 49) 0.000 338 696 181 750 169 6 × 2 = 0 + 0.000 677 392 363 500 339 2;
  • 50) 0.000 677 392 363 500 339 2 × 2 = 0 + 0.001 354 784 727 000 678 4;
  • 51) 0.001 354 784 727 000 678 4 × 2 = 0 + 0.002 709 569 454 001 356 8;
  • 52) 0.002 709 569 454 001 356 8 × 2 = 0 + 0.005 419 138 908 002 713 6;
  • 53) 0.005 419 138 908 002 713 6 × 2 = 0 + 0.010 838 277 816 005 427 2;
  • 54) 0.010 838 277 816 005 427 2 × 2 = 0 + 0.021 676 555 632 010 854 4;

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:


0.000 000 000 742 147 677 85(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 677 85(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:


0.000 000 000 742 147 677 85(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 -0.000 000 000 742 147 677 85 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