-0.000 000 000 742 147 688 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 688(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
-0.000 000 000 742 147 688(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:

|-0.000 000 000 742 147 688| = 0.000 000 000 742 147 688


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: 0.000 000 000 742 147 688.

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) 0.000 000 000 742 147 688 × 2 = 0 + 0.000 000 001 484 295 376;
  • 2) 0.000 000 001 484 295 376 × 2 = 0 + 0.000 000 002 968 590 752;
  • 3) 0.000 000 002 968 590 752 × 2 = 0 + 0.000 000 005 937 181 504;
  • 4) 0.000 000 005 937 181 504 × 2 = 0 + 0.000 000 011 874 363 008;
  • 5) 0.000 000 011 874 363 008 × 2 = 0 + 0.000 000 023 748 726 016;
  • 6) 0.000 000 023 748 726 016 × 2 = 0 + 0.000 000 047 497 452 032;
  • 7) 0.000 000 047 497 452 032 × 2 = 0 + 0.000 000 094 994 904 064;
  • 8) 0.000 000 094 994 904 064 × 2 = 0 + 0.000 000 189 989 808 128;
  • 9) 0.000 000 189 989 808 128 × 2 = 0 + 0.000 000 379 979 616 256;
  • 10) 0.000 000 379 979 616 256 × 2 = 0 + 0.000 000 759 959 232 512;
  • 11) 0.000 000 759 959 232 512 × 2 = 0 + 0.000 001 519 918 465 024;
  • 12) 0.000 001 519 918 465 024 × 2 = 0 + 0.000 003 039 836 930 048;
  • 13) 0.000 003 039 836 930 048 × 2 = 0 + 0.000 006 079 673 860 096;
  • 14) 0.000 006 079 673 860 096 × 2 = 0 + 0.000 012 159 347 720 192;
  • 15) 0.000 012 159 347 720 192 × 2 = 0 + 0.000 024 318 695 440 384;
  • 16) 0.000 024 318 695 440 384 × 2 = 0 + 0.000 048 637 390 880 768;
  • 17) 0.000 048 637 390 880 768 × 2 = 0 + 0.000 097 274 781 761 536;
  • 18) 0.000 097 274 781 761 536 × 2 = 0 + 0.000 194 549 563 523 072;
  • 19) 0.000 194 549 563 523 072 × 2 = 0 + 0.000 389 099 127 046 144;
  • 20) 0.000 389 099 127 046 144 × 2 = 0 + 0.000 778 198 254 092 288;
  • 21) 0.000 778 198 254 092 288 × 2 = 0 + 0.001 556 396 508 184 576;
  • 22) 0.001 556 396 508 184 576 × 2 = 0 + 0.003 112 793 016 369 152;
  • 23) 0.003 112 793 016 369 152 × 2 = 0 + 0.006 225 586 032 738 304;
  • 24) 0.006 225 586 032 738 304 × 2 = 0 + 0.012 451 172 065 476 608;
  • 25) 0.012 451 172 065 476 608 × 2 = 0 + 0.024 902 344 130 953 216;
  • 26) 0.024 902 344 130 953 216 × 2 = 0 + 0.049 804 688 261 906 432;
  • 27) 0.049 804 688 261 906 432 × 2 = 0 + 0.099 609 376 523 812 864;
  • 28) 0.099 609 376 523 812 864 × 2 = 0 + 0.199 218 753 047 625 728;
  • 29) 0.199 218 753 047 625 728 × 2 = 0 + 0.398 437 506 095 251 456;
  • 30) 0.398 437 506 095 251 456 × 2 = 0 + 0.796 875 012 190 502 912;
  • 31) 0.796 875 012 190 502 912 × 2 = 1 + 0.593 750 024 381 005 824;
  • 32) 0.593 750 024 381 005 824 × 2 = 1 + 0.187 500 048 762 011 648;
  • 33) 0.187 500 048 762 011 648 × 2 = 0 + 0.375 000 097 524 023 296;
  • 34) 0.375 000 097 524 023 296 × 2 = 0 + 0.750 000 195 048 046 592;
  • 35) 0.750 000 195 048 046 592 × 2 = 1 + 0.500 000 390 096 093 184;
  • 36) 0.500 000 390 096 093 184 × 2 = 1 + 0.000 000 780 192 186 368;
  • 37) 0.000 000 780 192 186 368 × 2 = 0 + 0.000 001 560 384 372 736;
  • 38) 0.000 001 560 384 372 736 × 2 = 0 + 0.000 003 120 768 745 472;
  • 39) 0.000 003 120 768 745 472 × 2 = 0 + 0.000 006 241 537 490 944;
  • 40) 0.000 006 241 537 490 944 × 2 = 0 + 0.000 012 483 074 981 888;
  • 41) 0.000 012 483 074 981 888 × 2 = 0 + 0.000 024 966 149 963 776;
  • 42) 0.000 024 966 149 963 776 × 2 = 0 + 0.000 049 932 299 927 552;
  • 43) 0.000 049 932 299 927 552 × 2 = 0 + 0.000 099 864 599 855 104;
  • 44) 0.000 099 864 599 855 104 × 2 = 0 + 0.000 199 729 199 710 208;
  • 45) 0.000 199 729 199 710 208 × 2 = 0 + 0.000 399 458 399 420 416;
  • 46) 0.000 399 458 399 420 416 × 2 = 0 + 0.000 798 916 798 840 832;
  • 47) 0.000 798 916 798 840 832 × 2 = 0 + 0.001 597 833 597 681 664;
  • 48) 0.001 597 833 597 681 664 × 2 = 0 + 0.003 195 667 195 363 328;
  • 49) 0.003 195 667 195 363 328 × 2 = 0 + 0.006 391 334 390 726 656;
  • 50) 0.006 391 334 390 726 656 × 2 = 0 + 0.012 782 668 781 453 312;
  • 51) 0.012 782 668 781 453 312 × 2 = 0 + 0.025 565 337 562 906 624;
  • 52) 0.025 565 337 562 906 624 × 2 = 0 + 0.051 130 675 125 813 248;
  • 53) 0.051 130 675 125 813 248 × 2 = 0 + 0.102 261 350 251 626 496;
  • 54) 0.102 261 350 251 626 496 × 2 = 0 + 0.204 522 700 503 252 992;

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:


0.000 000 000 742 147 688(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 688(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:


0.000 000 000 742 147 688(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 -0.000 000 000 742 147 688 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