-0.000 000 000 742 120 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 120 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
-0.000 000 000 742 120 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:

|-0.000 000 000 742 120 5| = 0.000 000 000 742 120 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: 0.000 000 000 742 120 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) 0.000 000 000 742 120 5 × 2 = 0 + 0.000 000 001 484 241;
  • 2) 0.000 000 001 484 241 × 2 = 0 + 0.000 000 002 968 482;
  • 3) 0.000 000 002 968 482 × 2 = 0 + 0.000 000 005 936 964;
  • 4) 0.000 000 005 936 964 × 2 = 0 + 0.000 000 011 873 928;
  • 5) 0.000 000 011 873 928 × 2 = 0 + 0.000 000 023 747 856;
  • 6) 0.000 000 023 747 856 × 2 = 0 + 0.000 000 047 495 712;
  • 7) 0.000 000 047 495 712 × 2 = 0 + 0.000 000 094 991 424;
  • 8) 0.000 000 094 991 424 × 2 = 0 + 0.000 000 189 982 848;
  • 9) 0.000 000 189 982 848 × 2 = 0 + 0.000 000 379 965 696;
  • 10) 0.000 000 379 965 696 × 2 = 0 + 0.000 000 759 931 392;
  • 11) 0.000 000 759 931 392 × 2 = 0 + 0.000 001 519 862 784;
  • 12) 0.000 001 519 862 784 × 2 = 0 + 0.000 003 039 725 568;
  • 13) 0.000 003 039 725 568 × 2 = 0 + 0.000 006 079 451 136;
  • 14) 0.000 006 079 451 136 × 2 = 0 + 0.000 012 158 902 272;
  • 15) 0.000 012 158 902 272 × 2 = 0 + 0.000 024 317 804 544;
  • 16) 0.000 024 317 804 544 × 2 = 0 + 0.000 048 635 609 088;
  • 17) 0.000 048 635 609 088 × 2 = 0 + 0.000 097 271 218 176;
  • 18) 0.000 097 271 218 176 × 2 = 0 + 0.000 194 542 436 352;
  • 19) 0.000 194 542 436 352 × 2 = 0 + 0.000 389 084 872 704;
  • 20) 0.000 389 084 872 704 × 2 = 0 + 0.000 778 169 745 408;
  • 21) 0.000 778 169 745 408 × 2 = 0 + 0.001 556 339 490 816;
  • 22) 0.001 556 339 490 816 × 2 = 0 + 0.003 112 678 981 632;
  • 23) 0.003 112 678 981 632 × 2 = 0 + 0.006 225 357 963 264;
  • 24) 0.006 225 357 963 264 × 2 = 0 + 0.012 450 715 926 528;
  • 25) 0.012 450 715 926 528 × 2 = 0 + 0.024 901 431 853 056;
  • 26) 0.024 901 431 853 056 × 2 = 0 + 0.049 802 863 706 112;
  • 27) 0.049 802 863 706 112 × 2 = 0 + 0.099 605 727 412 224;
  • 28) 0.099 605 727 412 224 × 2 = 0 + 0.199 211 454 824 448;
  • 29) 0.199 211 454 824 448 × 2 = 0 + 0.398 422 909 648 896;
  • 30) 0.398 422 909 648 896 × 2 = 0 + 0.796 845 819 297 792;
  • 31) 0.796 845 819 297 792 × 2 = 1 + 0.593 691 638 595 584;
  • 32) 0.593 691 638 595 584 × 2 = 1 + 0.187 383 277 191 168;
  • 33) 0.187 383 277 191 168 × 2 = 0 + 0.374 766 554 382 336;
  • 34) 0.374 766 554 382 336 × 2 = 0 + 0.749 533 108 764 672;
  • 35) 0.749 533 108 764 672 × 2 = 1 + 0.499 066 217 529 344;
  • 36) 0.499 066 217 529 344 × 2 = 0 + 0.998 132 435 058 688;
  • 37) 0.998 132 435 058 688 × 2 = 1 + 0.996 264 870 117 376;
  • 38) 0.996 264 870 117 376 × 2 = 1 + 0.992 529 740 234 752;
  • 39) 0.992 529 740 234 752 × 2 = 1 + 0.985 059 480 469 504;
  • 40) 0.985 059 480 469 504 × 2 = 1 + 0.970 118 960 939 008;
  • 41) 0.970 118 960 939 008 × 2 = 1 + 0.940 237 921 878 016;
  • 42) 0.940 237 921 878 016 × 2 = 1 + 0.880 475 843 756 032;
  • 43) 0.880 475 843 756 032 × 2 = 1 + 0.760 951 687 512 064;
  • 44) 0.760 951 687 512 064 × 2 = 1 + 0.521 903 375 024 128;
  • 45) 0.521 903 375 024 128 × 2 = 1 + 0.043 806 750 048 256;
  • 46) 0.043 806 750 048 256 × 2 = 0 + 0.087 613 500 096 512;
  • 47) 0.087 613 500 096 512 × 2 = 0 + 0.175 227 000 193 024;
  • 48) 0.175 227 000 193 024 × 2 = 0 + 0.350 454 000 386 048;
  • 49) 0.350 454 000 386 048 × 2 = 0 + 0.700 908 000 772 096;
  • 50) 0.700 908 000 772 096 × 2 = 1 + 0.401 816 001 544 192;
  • 51) 0.401 816 001 544 192 × 2 = 0 + 0.803 632 003 088 384;
  • 52) 0.803 632 003 088 384 × 2 = 1 + 0.607 264 006 176 768;
  • 53) 0.607 264 006 176 768 × 2 = 1 + 0.214 528 012 353 536;
  • 54) 0.214 528 012 353 536 × 2 = 0 + 0.429 056 024 707 072;

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:


0.000 000 000 742 120 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 0101 10(2)

6. Positive number before normalization:

0.000 000 000 742 120 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 0101 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:


0.000 000 000 742 120 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 0101 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 0101 10(2) × 20 =

1.1001 0111 1111 1100 0010 110(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 1100 0010 110


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 1110 0001 0110 =

100 1011 1111 1110 0001 0110


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 1110 0001 0110


Decimal number -0.000 000 000 742 120 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 0001 0110

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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