-0.000 000 000 742 147 646 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 646(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
-0.000 000 000 742 147 646(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:

|-0.000 000 000 742 147 646| = 0.000 000 000 742 147 646


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: 0.000 000 000 742 147 646.

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) 0.000 000 000 742 147 646 × 2 = 0 + 0.000 000 001 484 295 292;
  • 2) 0.000 000 001 484 295 292 × 2 = 0 + 0.000 000 002 968 590 584;
  • 3) 0.000 000 002 968 590 584 × 2 = 0 + 0.000 000 005 937 181 168;
  • 4) 0.000 000 005 937 181 168 × 2 = 0 + 0.000 000 011 874 362 336;
  • 5) 0.000 000 011 874 362 336 × 2 = 0 + 0.000 000 023 748 724 672;
  • 6) 0.000 000 023 748 724 672 × 2 = 0 + 0.000 000 047 497 449 344;
  • 7) 0.000 000 047 497 449 344 × 2 = 0 + 0.000 000 094 994 898 688;
  • 8) 0.000 000 094 994 898 688 × 2 = 0 + 0.000 000 189 989 797 376;
  • 9) 0.000 000 189 989 797 376 × 2 = 0 + 0.000 000 379 979 594 752;
  • 10) 0.000 000 379 979 594 752 × 2 = 0 + 0.000 000 759 959 189 504;
  • 11) 0.000 000 759 959 189 504 × 2 = 0 + 0.000 001 519 918 379 008;
  • 12) 0.000 001 519 918 379 008 × 2 = 0 + 0.000 003 039 836 758 016;
  • 13) 0.000 003 039 836 758 016 × 2 = 0 + 0.000 006 079 673 516 032;
  • 14) 0.000 006 079 673 516 032 × 2 = 0 + 0.000 012 159 347 032 064;
  • 15) 0.000 012 159 347 032 064 × 2 = 0 + 0.000 024 318 694 064 128;
  • 16) 0.000 024 318 694 064 128 × 2 = 0 + 0.000 048 637 388 128 256;
  • 17) 0.000 048 637 388 128 256 × 2 = 0 + 0.000 097 274 776 256 512;
  • 18) 0.000 097 274 776 256 512 × 2 = 0 + 0.000 194 549 552 513 024;
  • 19) 0.000 194 549 552 513 024 × 2 = 0 + 0.000 389 099 105 026 048;
  • 20) 0.000 389 099 105 026 048 × 2 = 0 + 0.000 778 198 210 052 096;
  • 21) 0.000 778 198 210 052 096 × 2 = 0 + 0.001 556 396 420 104 192;
  • 22) 0.001 556 396 420 104 192 × 2 = 0 + 0.003 112 792 840 208 384;
  • 23) 0.003 112 792 840 208 384 × 2 = 0 + 0.006 225 585 680 416 768;
  • 24) 0.006 225 585 680 416 768 × 2 = 0 + 0.012 451 171 360 833 536;
  • 25) 0.012 451 171 360 833 536 × 2 = 0 + 0.024 902 342 721 667 072;
  • 26) 0.024 902 342 721 667 072 × 2 = 0 + 0.049 804 685 443 334 144;
  • 27) 0.049 804 685 443 334 144 × 2 = 0 + 0.099 609 370 886 668 288;
  • 28) 0.099 609 370 886 668 288 × 2 = 0 + 0.199 218 741 773 336 576;
  • 29) 0.199 218 741 773 336 576 × 2 = 0 + 0.398 437 483 546 673 152;
  • 30) 0.398 437 483 546 673 152 × 2 = 0 + 0.796 874 967 093 346 304;
  • 31) 0.796 874 967 093 346 304 × 2 = 1 + 0.593 749 934 186 692 608;
  • 32) 0.593 749 934 186 692 608 × 2 = 1 + 0.187 499 868 373 385 216;
  • 33) 0.187 499 868 373 385 216 × 2 = 0 + 0.374 999 736 746 770 432;
  • 34) 0.374 999 736 746 770 432 × 2 = 0 + 0.749 999 473 493 540 864;
  • 35) 0.749 999 473 493 540 864 × 2 = 1 + 0.499 998 946 987 081 728;
  • 36) 0.499 998 946 987 081 728 × 2 = 0 + 0.999 997 893 974 163 456;
  • 37) 0.999 997 893 974 163 456 × 2 = 1 + 0.999 995 787 948 326 912;
  • 38) 0.999 995 787 948 326 912 × 2 = 1 + 0.999 991 575 896 653 824;
  • 39) 0.999 991 575 896 653 824 × 2 = 1 + 0.999 983 151 793 307 648;
  • 40) 0.999 983 151 793 307 648 × 2 = 1 + 0.999 966 303 586 615 296;
  • 41) 0.999 966 303 586 615 296 × 2 = 1 + 0.999 932 607 173 230 592;
  • 42) 0.999 932 607 173 230 592 × 2 = 1 + 0.999 865 214 346 461 184;
  • 43) 0.999 865 214 346 461 184 × 2 = 1 + 0.999 730 428 692 922 368;
  • 44) 0.999 730 428 692 922 368 × 2 = 1 + 0.999 460 857 385 844 736;
  • 45) 0.999 460 857 385 844 736 × 2 = 1 + 0.998 921 714 771 689 472;
  • 46) 0.998 921 714 771 689 472 × 2 = 1 + 0.997 843 429 543 378 944;
  • 47) 0.997 843 429 543 378 944 × 2 = 1 + 0.995 686 859 086 757 888;
  • 48) 0.995 686 859 086 757 888 × 2 = 1 + 0.991 373 718 173 515 776;
  • 49) 0.991 373 718 173 515 776 × 2 = 1 + 0.982 747 436 347 031 552;
  • 50) 0.982 747 436 347 031 552 × 2 = 1 + 0.965 494 872 694 063 104;
  • 51) 0.965 494 872 694 063 104 × 2 = 1 + 0.930 989 745 388 126 208;
  • 52) 0.930 989 745 388 126 208 × 2 = 1 + 0.861 979 490 776 252 416;
  • 53) 0.861 979 490 776 252 416 × 2 = 1 + 0.723 958 981 552 504 832;
  • 54) 0.723 958 981 552 504 832 × 2 = 1 + 0.447 917 963 105 009 664;

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:


0.000 000 000 742 147 646(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 646(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:


0.000 000 000 742 147 646(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 -0.000 000 000 742 147 646 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