-0.000 000 000 742 147 676 74 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 74(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 74(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 74| = 0.000 000 000 742 147 676 74


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 74.

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 74 × 2 = 0 + 0.000 000 001 484 295 353 48;
  • 2) 0.000 000 001 484 295 353 48 × 2 = 0 + 0.000 000 002 968 590 706 96;
  • 3) 0.000 000 002 968 590 706 96 × 2 = 0 + 0.000 000 005 937 181 413 92;
  • 4) 0.000 000 005 937 181 413 92 × 2 = 0 + 0.000 000 011 874 362 827 84;
  • 5) 0.000 000 011 874 362 827 84 × 2 = 0 + 0.000 000 023 748 725 655 68;
  • 6) 0.000 000 023 748 725 655 68 × 2 = 0 + 0.000 000 047 497 451 311 36;
  • 7) 0.000 000 047 497 451 311 36 × 2 = 0 + 0.000 000 094 994 902 622 72;
  • 8) 0.000 000 094 994 902 622 72 × 2 = 0 + 0.000 000 189 989 805 245 44;
  • 9) 0.000 000 189 989 805 245 44 × 2 = 0 + 0.000 000 379 979 610 490 88;
  • 10) 0.000 000 379 979 610 490 88 × 2 = 0 + 0.000 000 759 959 220 981 76;
  • 11) 0.000 000 759 959 220 981 76 × 2 = 0 + 0.000 001 519 918 441 963 52;
  • 12) 0.000 001 519 918 441 963 52 × 2 = 0 + 0.000 003 039 836 883 927 04;
  • 13) 0.000 003 039 836 883 927 04 × 2 = 0 + 0.000 006 079 673 767 854 08;
  • 14) 0.000 006 079 673 767 854 08 × 2 = 0 + 0.000 012 159 347 535 708 16;
  • 15) 0.000 012 159 347 535 708 16 × 2 = 0 + 0.000 024 318 695 071 416 32;
  • 16) 0.000 024 318 695 071 416 32 × 2 = 0 + 0.000 048 637 390 142 832 64;
  • 17) 0.000 048 637 390 142 832 64 × 2 = 0 + 0.000 097 274 780 285 665 28;
  • 18) 0.000 097 274 780 285 665 28 × 2 = 0 + 0.000 194 549 560 571 330 56;
  • 19) 0.000 194 549 560 571 330 56 × 2 = 0 + 0.000 389 099 121 142 661 12;
  • 20) 0.000 389 099 121 142 661 12 × 2 = 0 + 0.000 778 198 242 285 322 24;
  • 21) 0.000 778 198 242 285 322 24 × 2 = 0 + 0.001 556 396 484 570 644 48;
  • 22) 0.001 556 396 484 570 644 48 × 2 = 0 + 0.003 112 792 969 141 288 96;
  • 23) 0.003 112 792 969 141 288 96 × 2 = 0 + 0.006 225 585 938 282 577 92;
  • 24) 0.006 225 585 938 282 577 92 × 2 = 0 + 0.012 451 171 876 565 155 84;
  • 25) 0.012 451 171 876 565 155 84 × 2 = 0 + 0.024 902 343 753 130 311 68;
  • 26) 0.024 902 343 753 130 311 68 × 2 = 0 + 0.049 804 687 506 260 623 36;
  • 27) 0.049 804 687 506 260 623 36 × 2 = 0 + 0.099 609 375 012 521 246 72;
  • 28) 0.099 609 375 012 521 246 72 × 2 = 0 + 0.199 218 750 025 042 493 44;
  • 29) 0.199 218 750 025 042 493 44 × 2 = 0 + 0.398 437 500 050 084 986 88;
  • 30) 0.398 437 500 050 084 986 88 × 2 = 0 + 0.796 875 000 100 169 973 76;
  • 31) 0.796 875 000 100 169 973 76 × 2 = 1 + 0.593 750 000 200 339 947 52;
  • 32) 0.593 750 000 200 339 947 52 × 2 = 1 + 0.187 500 000 400 679 895 04;
  • 33) 0.187 500 000 400 679 895 04 × 2 = 0 + 0.375 000 000 801 359 790 08;
  • 34) 0.375 000 000 801 359 790 08 × 2 = 0 + 0.750 000 001 602 719 580 16;
  • 35) 0.750 000 001 602 719 580 16 × 2 = 1 + 0.500 000 003 205 439 160 32;
  • 36) 0.500 000 003 205 439 160 32 × 2 = 1 + 0.000 000 006 410 878 320 64;
  • 37) 0.000 000 006 410 878 320 64 × 2 = 0 + 0.000 000 012 821 756 641 28;
  • 38) 0.000 000 012 821 756 641 28 × 2 = 0 + 0.000 000 025 643 513 282 56;
  • 39) 0.000 000 025 643 513 282 56 × 2 = 0 + 0.000 000 051 287 026 565 12;
  • 40) 0.000 000 051 287 026 565 12 × 2 = 0 + 0.000 000 102 574 053 130 24;
  • 41) 0.000 000 102 574 053 130 24 × 2 = 0 + 0.000 000 205 148 106 260 48;
  • 42) 0.000 000 205 148 106 260 48 × 2 = 0 + 0.000 000 410 296 212 520 96;
  • 43) 0.000 000 410 296 212 520 96 × 2 = 0 + 0.000 000 820 592 425 041 92;
  • 44) 0.000 000 820 592 425 041 92 × 2 = 0 + 0.000 001 641 184 850 083 84;
  • 45) 0.000 001 641 184 850 083 84 × 2 = 0 + 0.000 003 282 369 700 167 68;
  • 46) 0.000 003 282 369 700 167 68 × 2 = 0 + 0.000 006 564 739 400 335 36;
  • 47) 0.000 006 564 739 400 335 36 × 2 = 0 + 0.000 013 129 478 800 670 72;
  • 48) 0.000 013 129 478 800 670 72 × 2 = 0 + 0.000 026 258 957 601 341 44;
  • 49) 0.000 026 258 957 601 341 44 × 2 = 0 + 0.000 052 517 915 202 682 88;
  • 50) 0.000 052 517 915 202 682 88 × 2 = 0 + 0.000 105 035 830 405 365 76;
  • 51) 0.000 105 035 830 405 365 76 × 2 = 0 + 0.000 210 071 660 810 731 52;
  • 52) 0.000 210 071 660 810 731 52 × 2 = 0 + 0.000 420 143 321 621 463 04;
  • 53) 0.000 420 143 321 621 463 04 × 2 = 0 + 0.000 840 286 643 242 926 08;
  • 54) 0.000 840 286 643 242 926 08 × 2 = 0 + 0.001 680 573 286 485 852 16;

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 74(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 74(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 74(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 74 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