-0.000 000 000 742 147 676 818 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 818(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
-0.000 000 000 742 147 676 818(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:

|-0.000 000 000 742 147 676 818| = 0.000 000 000 742 147 676 818


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: 0.000 000 000 742 147 676 818.

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) 0.000 000 000 742 147 676 818 × 2 = 0 + 0.000 000 001 484 295 353 636;
  • 2) 0.000 000 001 484 295 353 636 × 2 = 0 + 0.000 000 002 968 590 707 272;
  • 3) 0.000 000 002 968 590 707 272 × 2 = 0 + 0.000 000 005 937 181 414 544;
  • 4) 0.000 000 005 937 181 414 544 × 2 = 0 + 0.000 000 011 874 362 829 088;
  • 5) 0.000 000 011 874 362 829 088 × 2 = 0 + 0.000 000 023 748 725 658 176;
  • 6) 0.000 000 023 748 725 658 176 × 2 = 0 + 0.000 000 047 497 451 316 352;
  • 7) 0.000 000 047 497 451 316 352 × 2 = 0 + 0.000 000 094 994 902 632 704;
  • 8) 0.000 000 094 994 902 632 704 × 2 = 0 + 0.000 000 189 989 805 265 408;
  • 9) 0.000 000 189 989 805 265 408 × 2 = 0 + 0.000 000 379 979 610 530 816;
  • 10) 0.000 000 379 979 610 530 816 × 2 = 0 + 0.000 000 759 959 221 061 632;
  • 11) 0.000 000 759 959 221 061 632 × 2 = 0 + 0.000 001 519 918 442 123 264;
  • 12) 0.000 001 519 918 442 123 264 × 2 = 0 + 0.000 003 039 836 884 246 528;
  • 13) 0.000 003 039 836 884 246 528 × 2 = 0 + 0.000 006 079 673 768 493 056;
  • 14) 0.000 006 079 673 768 493 056 × 2 = 0 + 0.000 012 159 347 536 986 112;
  • 15) 0.000 012 159 347 536 986 112 × 2 = 0 + 0.000 024 318 695 073 972 224;
  • 16) 0.000 024 318 695 073 972 224 × 2 = 0 + 0.000 048 637 390 147 944 448;
  • 17) 0.000 048 637 390 147 944 448 × 2 = 0 + 0.000 097 274 780 295 888 896;
  • 18) 0.000 097 274 780 295 888 896 × 2 = 0 + 0.000 194 549 560 591 777 792;
  • 19) 0.000 194 549 560 591 777 792 × 2 = 0 + 0.000 389 099 121 183 555 584;
  • 20) 0.000 389 099 121 183 555 584 × 2 = 0 + 0.000 778 198 242 367 111 168;
  • 21) 0.000 778 198 242 367 111 168 × 2 = 0 + 0.001 556 396 484 734 222 336;
  • 22) 0.001 556 396 484 734 222 336 × 2 = 0 + 0.003 112 792 969 468 444 672;
  • 23) 0.003 112 792 969 468 444 672 × 2 = 0 + 0.006 225 585 938 936 889 344;
  • 24) 0.006 225 585 938 936 889 344 × 2 = 0 + 0.012 451 171 877 873 778 688;
  • 25) 0.012 451 171 877 873 778 688 × 2 = 0 + 0.024 902 343 755 747 557 376;
  • 26) 0.024 902 343 755 747 557 376 × 2 = 0 + 0.049 804 687 511 495 114 752;
  • 27) 0.049 804 687 511 495 114 752 × 2 = 0 + 0.099 609 375 022 990 229 504;
  • 28) 0.099 609 375 022 990 229 504 × 2 = 0 + 0.199 218 750 045 980 459 008;
  • 29) 0.199 218 750 045 980 459 008 × 2 = 0 + 0.398 437 500 091 960 918 016;
  • 30) 0.398 437 500 091 960 918 016 × 2 = 0 + 0.796 875 000 183 921 836 032;
  • 31) 0.796 875 000 183 921 836 032 × 2 = 1 + 0.593 750 000 367 843 672 064;
  • 32) 0.593 750 000 367 843 672 064 × 2 = 1 + 0.187 500 000 735 687 344 128;
  • 33) 0.187 500 000 735 687 344 128 × 2 = 0 + 0.375 000 001 471 374 688 256;
  • 34) 0.375 000 001 471 374 688 256 × 2 = 0 + 0.750 000 002 942 749 376 512;
  • 35) 0.750 000 002 942 749 376 512 × 2 = 1 + 0.500 000 005 885 498 753 024;
  • 36) 0.500 000 005 885 498 753 024 × 2 = 1 + 0.000 000 011 770 997 506 048;
  • 37) 0.000 000 011 770 997 506 048 × 2 = 0 + 0.000 000 023 541 995 012 096;
  • 38) 0.000 000 023 541 995 012 096 × 2 = 0 + 0.000 000 047 083 990 024 192;
  • 39) 0.000 000 047 083 990 024 192 × 2 = 0 + 0.000 000 094 167 980 048 384;
  • 40) 0.000 000 094 167 980 048 384 × 2 = 0 + 0.000 000 188 335 960 096 768;
  • 41) 0.000 000 188 335 960 096 768 × 2 = 0 + 0.000 000 376 671 920 193 536;
  • 42) 0.000 000 376 671 920 193 536 × 2 = 0 + 0.000 000 753 343 840 387 072;
  • 43) 0.000 000 753 343 840 387 072 × 2 = 0 + 0.000 001 506 687 680 774 144;
  • 44) 0.000 001 506 687 680 774 144 × 2 = 0 + 0.000 003 013 375 361 548 288;
  • 45) 0.000 003 013 375 361 548 288 × 2 = 0 + 0.000 006 026 750 723 096 576;
  • 46) 0.000 006 026 750 723 096 576 × 2 = 0 + 0.000 012 053 501 446 193 152;
  • 47) 0.000 012 053 501 446 193 152 × 2 = 0 + 0.000 024 107 002 892 386 304;
  • 48) 0.000 024 107 002 892 386 304 × 2 = 0 + 0.000 048 214 005 784 772 608;
  • 49) 0.000 048 214 005 784 772 608 × 2 = 0 + 0.000 096 428 011 569 545 216;
  • 50) 0.000 096 428 011 569 545 216 × 2 = 0 + 0.000 192 856 023 139 090 432;
  • 51) 0.000 192 856 023 139 090 432 × 2 = 0 + 0.000 385 712 046 278 180 864;
  • 52) 0.000 385 712 046 278 180 864 × 2 = 0 + 0.000 771 424 092 556 361 728;
  • 53) 0.000 771 424 092 556 361 728 × 2 = 0 + 0.001 542 848 185 112 723 456;
  • 54) 0.001 542 848 185 112 723 456 × 2 = 0 + 0.003 085 696 370 225 446 912;

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:


0.000 000 000 742 147 676 818(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 818(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:


0.000 000 000 742 147 676 818(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 -0.000 000 000 742 147 676 818 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 867 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 minutes

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