-0.000 000 000 742 147 559 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 559(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
-0.000 000 000 742 147 559(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:

|-0.000 000 000 742 147 559| = 0.000 000 000 742 147 559


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: 0.000 000 000 742 147 559.

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) 0.000 000 000 742 147 559 × 2 = 0 + 0.000 000 001 484 295 118;
  • 2) 0.000 000 001 484 295 118 × 2 = 0 + 0.000 000 002 968 590 236;
  • 3) 0.000 000 002 968 590 236 × 2 = 0 + 0.000 000 005 937 180 472;
  • 4) 0.000 000 005 937 180 472 × 2 = 0 + 0.000 000 011 874 360 944;
  • 5) 0.000 000 011 874 360 944 × 2 = 0 + 0.000 000 023 748 721 888;
  • 6) 0.000 000 023 748 721 888 × 2 = 0 + 0.000 000 047 497 443 776;
  • 7) 0.000 000 047 497 443 776 × 2 = 0 + 0.000 000 094 994 887 552;
  • 8) 0.000 000 094 994 887 552 × 2 = 0 + 0.000 000 189 989 775 104;
  • 9) 0.000 000 189 989 775 104 × 2 = 0 + 0.000 000 379 979 550 208;
  • 10) 0.000 000 379 979 550 208 × 2 = 0 + 0.000 000 759 959 100 416;
  • 11) 0.000 000 759 959 100 416 × 2 = 0 + 0.000 001 519 918 200 832;
  • 12) 0.000 001 519 918 200 832 × 2 = 0 + 0.000 003 039 836 401 664;
  • 13) 0.000 003 039 836 401 664 × 2 = 0 + 0.000 006 079 672 803 328;
  • 14) 0.000 006 079 672 803 328 × 2 = 0 + 0.000 012 159 345 606 656;
  • 15) 0.000 012 159 345 606 656 × 2 = 0 + 0.000 024 318 691 213 312;
  • 16) 0.000 024 318 691 213 312 × 2 = 0 + 0.000 048 637 382 426 624;
  • 17) 0.000 048 637 382 426 624 × 2 = 0 + 0.000 097 274 764 853 248;
  • 18) 0.000 097 274 764 853 248 × 2 = 0 + 0.000 194 549 529 706 496;
  • 19) 0.000 194 549 529 706 496 × 2 = 0 + 0.000 389 099 059 412 992;
  • 20) 0.000 389 099 059 412 992 × 2 = 0 + 0.000 778 198 118 825 984;
  • 21) 0.000 778 198 118 825 984 × 2 = 0 + 0.001 556 396 237 651 968;
  • 22) 0.001 556 396 237 651 968 × 2 = 0 + 0.003 112 792 475 303 936;
  • 23) 0.003 112 792 475 303 936 × 2 = 0 + 0.006 225 584 950 607 872;
  • 24) 0.006 225 584 950 607 872 × 2 = 0 + 0.012 451 169 901 215 744;
  • 25) 0.012 451 169 901 215 744 × 2 = 0 + 0.024 902 339 802 431 488;
  • 26) 0.024 902 339 802 431 488 × 2 = 0 + 0.049 804 679 604 862 976;
  • 27) 0.049 804 679 604 862 976 × 2 = 0 + 0.099 609 359 209 725 952;
  • 28) 0.099 609 359 209 725 952 × 2 = 0 + 0.199 218 718 419 451 904;
  • 29) 0.199 218 718 419 451 904 × 2 = 0 + 0.398 437 436 838 903 808;
  • 30) 0.398 437 436 838 903 808 × 2 = 0 + 0.796 874 873 677 807 616;
  • 31) 0.796 874 873 677 807 616 × 2 = 1 + 0.593 749 747 355 615 232;
  • 32) 0.593 749 747 355 615 232 × 2 = 1 + 0.187 499 494 711 230 464;
  • 33) 0.187 499 494 711 230 464 × 2 = 0 + 0.374 998 989 422 460 928;
  • 34) 0.374 998 989 422 460 928 × 2 = 0 + 0.749 997 978 844 921 856;
  • 35) 0.749 997 978 844 921 856 × 2 = 1 + 0.499 995 957 689 843 712;
  • 36) 0.499 995 957 689 843 712 × 2 = 0 + 0.999 991 915 379 687 424;
  • 37) 0.999 991 915 379 687 424 × 2 = 1 + 0.999 983 830 759 374 848;
  • 38) 0.999 983 830 759 374 848 × 2 = 1 + 0.999 967 661 518 749 696;
  • 39) 0.999 967 661 518 749 696 × 2 = 1 + 0.999 935 323 037 499 392;
  • 40) 0.999 935 323 037 499 392 × 2 = 1 + 0.999 870 646 074 998 784;
  • 41) 0.999 870 646 074 998 784 × 2 = 1 + 0.999 741 292 149 997 568;
  • 42) 0.999 741 292 149 997 568 × 2 = 1 + 0.999 482 584 299 995 136;
  • 43) 0.999 482 584 299 995 136 × 2 = 1 + 0.998 965 168 599 990 272;
  • 44) 0.998 965 168 599 990 272 × 2 = 1 + 0.997 930 337 199 980 544;
  • 45) 0.997 930 337 199 980 544 × 2 = 1 + 0.995 860 674 399 961 088;
  • 46) 0.995 860 674 399 961 088 × 2 = 1 + 0.991 721 348 799 922 176;
  • 47) 0.991 721 348 799 922 176 × 2 = 1 + 0.983 442 697 599 844 352;
  • 48) 0.983 442 697 599 844 352 × 2 = 1 + 0.966 885 395 199 688 704;
  • 49) 0.966 885 395 199 688 704 × 2 = 1 + 0.933 770 790 399 377 408;
  • 50) 0.933 770 790 399 377 408 × 2 = 1 + 0.867 541 580 798 754 816;
  • 51) 0.867 541 580 798 754 816 × 2 = 1 + 0.735 083 161 597 509 632;
  • 52) 0.735 083 161 597 509 632 × 2 = 1 + 0.470 166 323 195 019 264;
  • 53) 0.470 166 323 195 019 264 × 2 = 0 + 0.940 332 646 390 038 528;
  • 54) 0.940 332 646 390 038 528 × 2 = 1 + 0.880 665 292 780 077 056;

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:


0.000 000 000 742 147 559(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 559(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(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:


0.000 000 000 742 147 559(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) × 20 =

1.1001 0111 1111 1111 1111 101(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 101


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 1101 =

100 1011 1111 1111 1111 1101


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 1101


Decimal number -0.000 000 000 742 147 559 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1101

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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