-0.000 000 000 742 147 676 448 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 448(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
-0.000 000 000 742 147 676 448(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:

|-0.000 000 000 742 147 676 448| = 0.000 000 000 742 147 676 448


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: 0.000 000 000 742 147 676 448.

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) 0.000 000 000 742 147 676 448 × 2 = 0 + 0.000 000 001 484 295 352 896;
  • 2) 0.000 000 001 484 295 352 896 × 2 = 0 + 0.000 000 002 968 590 705 792;
  • 3) 0.000 000 002 968 590 705 792 × 2 = 0 + 0.000 000 005 937 181 411 584;
  • 4) 0.000 000 005 937 181 411 584 × 2 = 0 + 0.000 000 011 874 362 823 168;
  • 5) 0.000 000 011 874 362 823 168 × 2 = 0 + 0.000 000 023 748 725 646 336;
  • 6) 0.000 000 023 748 725 646 336 × 2 = 0 + 0.000 000 047 497 451 292 672;
  • 7) 0.000 000 047 497 451 292 672 × 2 = 0 + 0.000 000 094 994 902 585 344;
  • 8) 0.000 000 094 994 902 585 344 × 2 = 0 + 0.000 000 189 989 805 170 688;
  • 9) 0.000 000 189 989 805 170 688 × 2 = 0 + 0.000 000 379 979 610 341 376;
  • 10) 0.000 000 379 979 610 341 376 × 2 = 0 + 0.000 000 759 959 220 682 752;
  • 11) 0.000 000 759 959 220 682 752 × 2 = 0 + 0.000 001 519 918 441 365 504;
  • 12) 0.000 001 519 918 441 365 504 × 2 = 0 + 0.000 003 039 836 882 731 008;
  • 13) 0.000 003 039 836 882 731 008 × 2 = 0 + 0.000 006 079 673 765 462 016;
  • 14) 0.000 006 079 673 765 462 016 × 2 = 0 + 0.000 012 159 347 530 924 032;
  • 15) 0.000 012 159 347 530 924 032 × 2 = 0 + 0.000 024 318 695 061 848 064;
  • 16) 0.000 024 318 695 061 848 064 × 2 = 0 + 0.000 048 637 390 123 696 128;
  • 17) 0.000 048 637 390 123 696 128 × 2 = 0 + 0.000 097 274 780 247 392 256;
  • 18) 0.000 097 274 780 247 392 256 × 2 = 0 + 0.000 194 549 560 494 784 512;
  • 19) 0.000 194 549 560 494 784 512 × 2 = 0 + 0.000 389 099 120 989 569 024;
  • 20) 0.000 389 099 120 989 569 024 × 2 = 0 + 0.000 778 198 241 979 138 048;
  • 21) 0.000 778 198 241 979 138 048 × 2 = 0 + 0.001 556 396 483 958 276 096;
  • 22) 0.001 556 396 483 958 276 096 × 2 = 0 + 0.003 112 792 967 916 552 192;
  • 23) 0.003 112 792 967 916 552 192 × 2 = 0 + 0.006 225 585 935 833 104 384;
  • 24) 0.006 225 585 935 833 104 384 × 2 = 0 + 0.012 451 171 871 666 208 768;
  • 25) 0.012 451 171 871 666 208 768 × 2 = 0 + 0.024 902 343 743 332 417 536;
  • 26) 0.024 902 343 743 332 417 536 × 2 = 0 + 0.049 804 687 486 664 835 072;
  • 27) 0.049 804 687 486 664 835 072 × 2 = 0 + 0.099 609 374 973 329 670 144;
  • 28) 0.099 609 374 973 329 670 144 × 2 = 0 + 0.199 218 749 946 659 340 288;
  • 29) 0.199 218 749 946 659 340 288 × 2 = 0 + 0.398 437 499 893 318 680 576;
  • 30) 0.398 437 499 893 318 680 576 × 2 = 0 + 0.796 874 999 786 637 361 152;
  • 31) 0.796 874 999 786 637 361 152 × 2 = 1 + 0.593 749 999 573 274 722 304;
  • 32) 0.593 749 999 573 274 722 304 × 2 = 1 + 0.187 499 999 146 549 444 608;
  • 33) 0.187 499 999 146 549 444 608 × 2 = 0 + 0.374 999 998 293 098 889 216;
  • 34) 0.374 999 998 293 098 889 216 × 2 = 0 + 0.749 999 996 586 197 778 432;
  • 35) 0.749 999 996 586 197 778 432 × 2 = 1 + 0.499 999 993 172 395 556 864;
  • 36) 0.499 999 993 172 395 556 864 × 2 = 0 + 0.999 999 986 344 791 113 728;
  • 37) 0.999 999 986 344 791 113 728 × 2 = 1 + 0.999 999 972 689 582 227 456;
  • 38) 0.999 999 972 689 582 227 456 × 2 = 1 + 0.999 999 945 379 164 454 912;
  • 39) 0.999 999 945 379 164 454 912 × 2 = 1 + 0.999 999 890 758 328 909 824;
  • 40) 0.999 999 890 758 328 909 824 × 2 = 1 + 0.999 999 781 516 657 819 648;
  • 41) 0.999 999 781 516 657 819 648 × 2 = 1 + 0.999 999 563 033 315 639 296;
  • 42) 0.999 999 563 033 315 639 296 × 2 = 1 + 0.999 999 126 066 631 278 592;
  • 43) 0.999 999 126 066 631 278 592 × 2 = 1 + 0.999 998 252 133 262 557 184;
  • 44) 0.999 998 252 133 262 557 184 × 2 = 1 + 0.999 996 504 266 525 114 368;
  • 45) 0.999 996 504 266 525 114 368 × 2 = 1 + 0.999 993 008 533 050 228 736;
  • 46) 0.999 993 008 533 050 228 736 × 2 = 1 + 0.999 986 017 066 100 457 472;
  • 47) 0.999 986 017 066 100 457 472 × 2 = 1 + 0.999 972 034 132 200 914 944;
  • 48) 0.999 972 034 132 200 914 944 × 2 = 1 + 0.999 944 068 264 401 829 888;
  • 49) 0.999 944 068 264 401 829 888 × 2 = 1 + 0.999 888 136 528 803 659 776;
  • 50) 0.999 888 136 528 803 659 776 × 2 = 1 + 0.999 776 273 057 607 319 552;
  • 51) 0.999 776 273 057 607 319 552 × 2 = 1 + 0.999 552 546 115 214 639 104;
  • 52) 0.999 552 546 115 214 639 104 × 2 = 1 + 0.999 105 092 230 429 278 208;
  • 53) 0.999 105 092 230 429 278 208 × 2 = 1 + 0.998 210 184 460 858 556 416;
  • 54) 0.998 210 184 460 858 556 416 × 2 = 1 + 0.996 420 368 921 717 112 832;

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:


0.000 000 000 742 147 676 448(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 448(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:


0.000 000 000 742 147 676 448(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 -0.000 000 000 742 147 676 448 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