-0.000 000 000 742 147 537 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 537(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
-0.000 000 000 742 147 537(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:

|-0.000 000 000 742 147 537| = 0.000 000 000 742 147 537


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: 0.000 000 000 742 147 537.

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) 0.000 000 000 742 147 537 × 2 = 0 + 0.000 000 001 484 295 074;
  • 2) 0.000 000 001 484 295 074 × 2 = 0 + 0.000 000 002 968 590 148;
  • 3) 0.000 000 002 968 590 148 × 2 = 0 + 0.000 000 005 937 180 296;
  • 4) 0.000 000 005 937 180 296 × 2 = 0 + 0.000 000 011 874 360 592;
  • 5) 0.000 000 011 874 360 592 × 2 = 0 + 0.000 000 023 748 721 184;
  • 6) 0.000 000 023 748 721 184 × 2 = 0 + 0.000 000 047 497 442 368;
  • 7) 0.000 000 047 497 442 368 × 2 = 0 + 0.000 000 094 994 884 736;
  • 8) 0.000 000 094 994 884 736 × 2 = 0 + 0.000 000 189 989 769 472;
  • 9) 0.000 000 189 989 769 472 × 2 = 0 + 0.000 000 379 979 538 944;
  • 10) 0.000 000 379 979 538 944 × 2 = 0 + 0.000 000 759 959 077 888;
  • 11) 0.000 000 759 959 077 888 × 2 = 0 + 0.000 001 519 918 155 776;
  • 12) 0.000 001 519 918 155 776 × 2 = 0 + 0.000 003 039 836 311 552;
  • 13) 0.000 003 039 836 311 552 × 2 = 0 + 0.000 006 079 672 623 104;
  • 14) 0.000 006 079 672 623 104 × 2 = 0 + 0.000 012 159 345 246 208;
  • 15) 0.000 012 159 345 246 208 × 2 = 0 + 0.000 024 318 690 492 416;
  • 16) 0.000 024 318 690 492 416 × 2 = 0 + 0.000 048 637 380 984 832;
  • 17) 0.000 048 637 380 984 832 × 2 = 0 + 0.000 097 274 761 969 664;
  • 18) 0.000 097 274 761 969 664 × 2 = 0 + 0.000 194 549 523 939 328;
  • 19) 0.000 194 549 523 939 328 × 2 = 0 + 0.000 389 099 047 878 656;
  • 20) 0.000 389 099 047 878 656 × 2 = 0 + 0.000 778 198 095 757 312;
  • 21) 0.000 778 198 095 757 312 × 2 = 0 + 0.001 556 396 191 514 624;
  • 22) 0.001 556 396 191 514 624 × 2 = 0 + 0.003 112 792 383 029 248;
  • 23) 0.003 112 792 383 029 248 × 2 = 0 + 0.006 225 584 766 058 496;
  • 24) 0.006 225 584 766 058 496 × 2 = 0 + 0.012 451 169 532 116 992;
  • 25) 0.012 451 169 532 116 992 × 2 = 0 + 0.024 902 339 064 233 984;
  • 26) 0.024 902 339 064 233 984 × 2 = 0 + 0.049 804 678 128 467 968;
  • 27) 0.049 804 678 128 467 968 × 2 = 0 + 0.099 609 356 256 935 936;
  • 28) 0.099 609 356 256 935 936 × 2 = 0 + 0.199 218 712 513 871 872;
  • 29) 0.199 218 712 513 871 872 × 2 = 0 + 0.398 437 425 027 743 744;
  • 30) 0.398 437 425 027 743 744 × 2 = 0 + 0.796 874 850 055 487 488;
  • 31) 0.796 874 850 055 487 488 × 2 = 1 + 0.593 749 700 110 974 976;
  • 32) 0.593 749 700 110 974 976 × 2 = 1 + 0.187 499 400 221 949 952;
  • 33) 0.187 499 400 221 949 952 × 2 = 0 + 0.374 998 800 443 899 904;
  • 34) 0.374 998 800 443 899 904 × 2 = 0 + 0.749 997 600 887 799 808;
  • 35) 0.749 997 600 887 799 808 × 2 = 1 + 0.499 995 201 775 599 616;
  • 36) 0.499 995 201 775 599 616 × 2 = 0 + 0.999 990 403 551 199 232;
  • 37) 0.999 990 403 551 199 232 × 2 = 1 + 0.999 980 807 102 398 464;
  • 38) 0.999 980 807 102 398 464 × 2 = 1 + 0.999 961 614 204 796 928;
  • 39) 0.999 961 614 204 796 928 × 2 = 1 + 0.999 923 228 409 593 856;
  • 40) 0.999 923 228 409 593 856 × 2 = 1 + 0.999 846 456 819 187 712;
  • 41) 0.999 846 456 819 187 712 × 2 = 1 + 0.999 692 913 638 375 424;
  • 42) 0.999 692 913 638 375 424 × 2 = 1 + 0.999 385 827 276 750 848;
  • 43) 0.999 385 827 276 750 848 × 2 = 1 + 0.998 771 654 553 501 696;
  • 44) 0.998 771 654 553 501 696 × 2 = 1 + 0.997 543 309 107 003 392;
  • 45) 0.997 543 309 107 003 392 × 2 = 1 + 0.995 086 618 214 006 784;
  • 46) 0.995 086 618 214 006 784 × 2 = 1 + 0.990 173 236 428 013 568;
  • 47) 0.990 173 236 428 013 568 × 2 = 1 + 0.980 346 472 856 027 136;
  • 48) 0.980 346 472 856 027 136 × 2 = 1 + 0.960 692 945 712 054 272;
  • 49) 0.960 692 945 712 054 272 × 2 = 1 + 0.921 385 891 424 108 544;
  • 50) 0.921 385 891 424 108 544 × 2 = 1 + 0.842 771 782 848 217 088;
  • 51) 0.842 771 782 848 217 088 × 2 = 1 + 0.685 543 565 696 434 176;
  • 52) 0.685 543 565 696 434 176 × 2 = 1 + 0.371 087 131 392 868 352;
  • 53) 0.371 087 131 392 868 352 × 2 = 0 + 0.742 174 262 785 736 704;
  • 54) 0.742 174 262 785 736 704 × 2 = 1 + 0.484 348 525 571 473 408;

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:


0.000 000 000 742 147 537(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 537(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:


0.000 000 000 742 147 537(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 -0.000 000 000 742 147 537 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