-0.000 000 000 742 147 674 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674(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
-0.000 000 000 742 147 674(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:

|-0.000 000 000 742 147 674| = 0.000 000 000 742 147 674


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: 0.000 000 000 742 147 674.

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) 0.000 000 000 742 147 674 × 2 = 0 + 0.000 000 001 484 295 348;
  • 2) 0.000 000 001 484 295 348 × 2 = 0 + 0.000 000 002 968 590 696;
  • 3) 0.000 000 002 968 590 696 × 2 = 0 + 0.000 000 005 937 181 392;
  • 4) 0.000 000 005 937 181 392 × 2 = 0 + 0.000 000 011 874 362 784;
  • 5) 0.000 000 011 874 362 784 × 2 = 0 + 0.000 000 023 748 725 568;
  • 6) 0.000 000 023 748 725 568 × 2 = 0 + 0.000 000 047 497 451 136;
  • 7) 0.000 000 047 497 451 136 × 2 = 0 + 0.000 000 094 994 902 272;
  • 8) 0.000 000 094 994 902 272 × 2 = 0 + 0.000 000 189 989 804 544;
  • 9) 0.000 000 189 989 804 544 × 2 = 0 + 0.000 000 379 979 609 088;
  • 10) 0.000 000 379 979 609 088 × 2 = 0 + 0.000 000 759 959 218 176;
  • 11) 0.000 000 759 959 218 176 × 2 = 0 + 0.000 001 519 918 436 352;
  • 12) 0.000 001 519 918 436 352 × 2 = 0 + 0.000 003 039 836 872 704;
  • 13) 0.000 003 039 836 872 704 × 2 = 0 + 0.000 006 079 673 745 408;
  • 14) 0.000 006 079 673 745 408 × 2 = 0 + 0.000 012 159 347 490 816;
  • 15) 0.000 012 159 347 490 816 × 2 = 0 + 0.000 024 318 694 981 632;
  • 16) 0.000 024 318 694 981 632 × 2 = 0 + 0.000 048 637 389 963 264;
  • 17) 0.000 048 637 389 963 264 × 2 = 0 + 0.000 097 274 779 926 528;
  • 18) 0.000 097 274 779 926 528 × 2 = 0 + 0.000 194 549 559 853 056;
  • 19) 0.000 194 549 559 853 056 × 2 = 0 + 0.000 389 099 119 706 112;
  • 20) 0.000 389 099 119 706 112 × 2 = 0 + 0.000 778 198 239 412 224;
  • 21) 0.000 778 198 239 412 224 × 2 = 0 + 0.001 556 396 478 824 448;
  • 22) 0.001 556 396 478 824 448 × 2 = 0 + 0.003 112 792 957 648 896;
  • 23) 0.003 112 792 957 648 896 × 2 = 0 + 0.006 225 585 915 297 792;
  • 24) 0.006 225 585 915 297 792 × 2 = 0 + 0.012 451 171 830 595 584;
  • 25) 0.012 451 171 830 595 584 × 2 = 0 + 0.024 902 343 661 191 168;
  • 26) 0.024 902 343 661 191 168 × 2 = 0 + 0.049 804 687 322 382 336;
  • 27) 0.049 804 687 322 382 336 × 2 = 0 + 0.099 609 374 644 764 672;
  • 28) 0.099 609 374 644 764 672 × 2 = 0 + 0.199 218 749 289 529 344;
  • 29) 0.199 218 749 289 529 344 × 2 = 0 + 0.398 437 498 579 058 688;
  • 30) 0.398 437 498 579 058 688 × 2 = 0 + 0.796 874 997 158 117 376;
  • 31) 0.796 874 997 158 117 376 × 2 = 1 + 0.593 749 994 316 234 752;
  • 32) 0.593 749 994 316 234 752 × 2 = 1 + 0.187 499 988 632 469 504;
  • 33) 0.187 499 988 632 469 504 × 2 = 0 + 0.374 999 977 264 939 008;
  • 34) 0.374 999 977 264 939 008 × 2 = 0 + 0.749 999 954 529 878 016;
  • 35) 0.749 999 954 529 878 016 × 2 = 1 + 0.499 999 909 059 756 032;
  • 36) 0.499 999 909 059 756 032 × 2 = 0 + 0.999 999 818 119 512 064;
  • 37) 0.999 999 818 119 512 064 × 2 = 1 + 0.999 999 636 239 024 128;
  • 38) 0.999 999 636 239 024 128 × 2 = 1 + 0.999 999 272 478 048 256;
  • 39) 0.999 999 272 478 048 256 × 2 = 1 + 0.999 998 544 956 096 512;
  • 40) 0.999 998 544 956 096 512 × 2 = 1 + 0.999 997 089 912 193 024;
  • 41) 0.999 997 089 912 193 024 × 2 = 1 + 0.999 994 179 824 386 048;
  • 42) 0.999 994 179 824 386 048 × 2 = 1 + 0.999 988 359 648 772 096;
  • 43) 0.999 988 359 648 772 096 × 2 = 1 + 0.999 976 719 297 544 192;
  • 44) 0.999 976 719 297 544 192 × 2 = 1 + 0.999 953 438 595 088 384;
  • 45) 0.999 953 438 595 088 384 × 2 = 1 + 0.999 906 877 190 176 768;
  • 46) 0.999 906 877 190 176 768 × 2 = 1 + 0.999 813 754 380 353 536;
  • 47) 0.999 813 754 380 353 536 × 2 = 1 + 0.999 627 508 760 707 072;
  • 48) 0.999 627 508 760 707 072 × 2 = 1 + 0.999 255 017 521 414 144;
  • 49) 0.999 255 017 521 414 144 × 2 = 1 + 0.998 510 035 042 828 288;
  • 50) 0.998 510 035 042 828 288 × 2 = 1 + 0.997 020 070 085 656 576;
  • 51) 0.997 020 070 085 656 576 × 2 = 1 + 0.994 040 140 171 313 152;
  • 52) 0.994 040 140 171 313 152 × 2 = 1 + 0.988 080 280 342 626 304;
  • 53) 0.988 080 280 342 626 304 × 2 = 1 + 0.976 160 560 685 252 608;
  • 54) 0.976 160 560 685 252 608 × 2 = 1 + 0.952 321 121 370 505 216;

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:


0.000 000 000 742 147 674(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 674(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:


0.000 000 000 742 147 674(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 -0.000 000 000 742 147 674 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