-0.000 000 000 742 149 57 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 149 57(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
-0.000 000 000 742 149 57(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:

|-0.000 000 000 742 149 57| = 0.000 000 000 742 149 57


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: 0.000 000 000 742 149 57.

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) 0.000 000 000 742 149 57 × 2 = 0 + 0.000 000 001 484 299 14;
  • 2) 0.000 000 001 484 299 14 × 2 = 0 + 0.000 000 002 968 598 28;
  • 3) 0.000 000 002 968 598 28 × 2 = 0 + 0.000 000 005 937 196 56;
  • 4) 0.000 000 005 937 196 56 × 2 = 0 + 0.000 000 011 874 393 12;
  • 5) 0.000 000 011 874 393 12 × 2 = 0 + 0.000 000 023 748 786 24;
  • 6) 0.000 000 023 748 786 24 × 2 = 0 + 0.000 000 047 497 572 48;
  • 7) 0.000 000 047 497 572 48 × 2 = 0 + 0.000 000 094 995 144 96;
  • 8) 0.000 000 094 995 144 96 × 2 = 0 + 0.000 000 189 990 289 92;
  • 9) 0.000 000 189 990 289 92 × 2 = 0 + 0.000 000 379 980 579 84;
  • 10) 0.000 000 379 980 579 84 × 2 = 0 + 0.000 000 759 961 159 68;
  • 11) 0.000 000 759 961 159 68 × 2 = 0 + 0.000 001 519 922 319 36;
  • 12) 0.000 001 519 922 319 36 × 2 = 0 + 0.000 003 039 844 638 72;
  • 13) 0.000 003 039 844 638 72 × 2 = 0 + 0.000 006 079 689 277 44;
  • 14) 0.000 006 079 689 277 44 × 2 = 0 + 0.000 012 159 378 554 88;
  • 15) 0.000 012 159 378 554 88 × 2 = 0 + 0.000 024 318 757 109 76;
  • 16) 0.000 024 318 757 109 76 × 2 = 0 + 0.000 048 637 514 219 52;
  • 17) 0.000 048 637 514 219 52 × 2 = 0 + 0.000 097 275 028 439 04;
  • 18) 0.000 097 275 028 439 04 × 2 = 0 + 0.000 194 550 056 878 08;
  • 19) 0.000 194 550 056 878 08 × 2 = 0 + 0.000 389 100 113 756 16;
  • 20) 0.000 389 100 113 756 16 × 2 = 0 + 0.000 778 200 227 512 32;
  • 21) 0.000 778 200 227 512 32 × 2 = 0 + 0.001 556 400 455 024 64;
  • 22) 0.001 556 400 455 024 64 × 2 = 0 + 0.003 112 800 910 049 28;
  • 23) 0.003 112 800 910 049 28 × 2 = 0 + 0.006 225 601 820 098 56;
  • 24) 0.006 225 601 820 098 56 × 2 = 0 + 0.012 451 203 640 197 12;
  • 25) 0.012 451 203 640 197 12 × 2 = 0 + 0.024 902 407 280 394 24;
  • 26) 0.024 902 407 280 394 24 × 2 = 0 + 0.049 804 814 560 788 48;
  • 27) 0.049 804 814 560 788 48 × 2 = 0 + 0.099 609 629 121 576 96;
  • 28) 0.099 609 629 121 576 96 × 2 = 0 + 0.199 219 258 243 153 92;
  • 29) 0.199 219 258 243 153 92 × 2 = 0 + 0.398 438 516 486 307 84;
  • 30) 0.398 438 516 486 307 84 × 2 = 0 + 0.796 877 032 972 615 68;
  • 31) 0.796 877 032 972 615 68 × 2 = 1 + 0.593 754 065 945 231 36;
  • 32) 0.593 754 065 945 231 36 × 2 = 1 + 0.187 508 131 890 462 72;
  • 33) 0.187 508 131 890 462 72 × 2 = 0 + 0.375 016 263 780 925 44;
  • 34) 0.375 016 263 780 925 44 × 2 = 0 + 0.750 032 527 561 850 88;
  • 35) 0.750 032 527 561 850 88 × 2 = 1 + 0.500 065 055 123 701 76;
  • 36) 0.500 065 055 123 701 76 × 2 = 1 + 0.000 130 110 247 403 52;
  • 37) 0.000 130 110 247 403 52 × 2 = 0 + 0.000 260 220 494 807 04;
  • 38) 0.000 260 220 494 807 04 × 2 = 0 + 0.000 520 440 989 614 08;
  • 39) 0.000 520 440 989 614 08 × 2 = 0 + 0.001 040 881 979 228 16;
  • 40) 0.001 040 881 979 228 16 × 2 = 0 + 0.002 081 763 958 456 32;
  • 41) 0.002 081 763 958 456 32 × 2 = 0 + 0.004 163 527 916 912 64;
  • 42) 0.004 163 527 916 912 64 × 2 = 0 + 0.008 327 055 833 825 28;
  • 43) 0.008 327 055 833 825 28 × 2 = 0 + 0.016 654 111 667 650 56;
  • 44) 0.016 654 111 667 650 56 × 2 = 0 + 0.033 308 223 335 301 12;
  • 45) 0.033 308 223 335 301 12 × 2 = 0 + 0.066 616 446 670 602 24;
  • 46) 0.066 616 446 670 602 24 × 2 = 0 + 0.133 232 893 341 204 48;
  • 47) 0.133 232 893 341 204 48 × 2 = 0 + 0.266 465 786 682 408 96;
  • 48) 0.266 465 786 682 408 96 × 2 = 0 + 0.532 931 573 364 817 92;
  • 49) 0.532 931 573 364 817 92 × 2 = 1 + 0.065 863 146 729 635 84;
  • 50) 0.065 863 146 729 635 84 × 2 = 0 + 0.131 726 293 459 271 68;
  • 51) 0.131 726 293 459 271 68 × 2 = 0 + 0.263 452 586 918 543 36;
  • 52) 0.263 452 586 918 543 36 × 2 = 0 + 0.526 905 173 837 086 72;
  • 53) 0.526 905 173 837 086 72 × 2 = 1 + 0.053 810 347 674 173 44;
  • 54) 0.053 810 347 674 173 44 × 2 = 0 + 0.107 620 695 348 346 88;

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:


0.000 000 000 742 149 57(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1000 10(2)

6. Positive number before normalization:

0.000 000 000 742 149 57(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1000 10(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:


0.000 000 000 742 149 57(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1000 10(2) × 20 =

1.1001 1000 0000 0000 0100 010(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 0100 010


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 0010 0010 =

100 1100 0000 0000 0010 0010


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 0010 0010


Decimal number -0.000 000 000 742 149 57 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0010 0010

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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