-0.000 000 000 742 147 676 646 753 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 753(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
-0.000 000 000 742 147 676 646 753(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:

|-0.000 000 000 742 147 676 646 753| = 0.000 000 000 742 147 676 646 753


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: 0.000 000 000 742 147 676 646 753.

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) 0.000 000 000 742 147 676 646 753 × 2 = 0 + 0.000 000 001 484 295 353 293 506;
  • 2) 0.000 000 001 484 295 353 293 506 × 2 = 0 + 0.000 000 002 968 590 706 587 012;
  • 3) 0.000 000 002 968 590 706 587 012 × 2 = 0 + 0.000 000 005 937 181 413 174 024;
  • 4) 0.000 000 005 937 181 413 174 024 × 2 = 0 + 0.000 000 011 874 362 826 348 048;
  • 5) 0.000 000 011 874 362 826 348 048 × 2 = 0 + 0.000 000 023 748 725 652 696 096;
  • 6) 0.000 000 023 748 725 652 696 096 × 2 = 0 + 0.000 000 047 497 451 305 392 192;
  • 7) 0.000 000 047 497 451 305 392 192 × 2 = 0 + 0.000 000 094 994 902 610 784 384;
  • 8) 0.000 000 094 994 902 610 784 384 × 2 = 0 + 0.000 000 189 989 805 221 568 768;
  • 9) 0.000 000 189 989 805 221 568 768 × 2 = 0 + 0.000 000 379 979 610 443 137 536;
  • 10) 0.000 000 379 979 610 443 137 536 × 2 = 0 + 0.000 000 759 959 220 886 275 072;
  • 11) 0.000 000 759 959 220 886 275 072 × 2 = 0 + 0.000 001 519 918 441 772 550 144;
  • 12) 0.000 001 519 918 441 772 550 144 × 2 = 0 + 0.000 003 039 836 883 545 100 288;
  • 13) 0.000 003 039 836 883 545 100 288 × 2 = 0 + 0.000 006 079 673 767 090 200 576;
  • 14) 0.000 006 079 673 767 090 200 576 × 2 = 0 + 0.000 012 159 347 534 180 401 152;
  • 15) 0.000 012 159 347 534 180 401 152 × 2 = 0 + 0.000 024 318 695 068 360 802 304;
  • 16) 0.000 024 318 695 068 360 802 304 × 2 = 0 + 0.000 048 637 390 136 721 604 608;
  • 17) 0.000 048 637 390 136 721 604 608 × 2 = 0 + 0.000 097 274 780 273 443 209 216;
  • 18) 0.000 097 274 780 273 443 209 216 × 2 = 0 + 0.000 194 549 560 546 886 418 432;
  • 19) 0.000 194 549 560 546 886 418 432 × 2 = 0 + 0.000 389 099 121 093 772 836 864;
  • 20) 0.000 389 099 121 093 772 836 864 × 2 = 0 + 0.000 778 198 242 187 545 673 728;
  • 21) 0.000 778 198 242 187 545 673 728 × 2 = 0 + 0.001 556 396 484 375 091 347 456;
  • 22) 0.001 556 396 484 375 091 347 456 × 2 = 0 + 0.003 112 792 968 750 182 694 912;
  • 23) 0.003 112 792 968 750 182 694 912 × 2 = 0 + 0.006 225 585 937 500 365 389 824;
  • 24) 0.006 225 585 937 500 365 389 824 × 2 = 0 + 0.012 451 171 875 000 730 779 648;
  • 25) 0.012 451 171 875 000 730 779 648 × 2 = 0 + 0.024 902 343 750 001 461 559 296;
  • 26) 0.024 902 343 750 001 461 559 296 × 2 = 0 + 0.049 804 687 500 002 923 118 592;
  • 27) 0.049 804 687 500 002 923 118 592 × 2 = 0 + 0.099 609 375 000 005 846 237 184;
  • 28) 0.099 609 375 000 005 846 237 184 × 2 = 0 + 0.199 218 750 000 011 692 474 368;
  • 29) 0.199 218 750 000 011 692 474 368 × 2 = 0 + 0.398 437 500 000 023 384 948 736;
  • 30) 0.398 437 500 000 023 384 948 736 × 2 = 0 + 0.796 875 000 000 046 769 897 472;
  • 31) 0.796 875 000 000 046 769 897 472 × 2 = 1 + 0.593 750 000 000 093 539 794 944;
  • 32) 0.593 750 000 000 093 539 794 944 × 2 = 1 + 0.187 500 000 000 187 079 589 888;
  • 33) 0.187 500 000 000 187 079 589 888 × 2 = 0 + 0.375 000 000 000 374 159 179 776;
  • 34) 0.375 000 000 000 374 159 179 776 × 2 = 0 + 0.750 000 000 000 748 318 359 552;
  • 35) 0.750 000 000 000 748 318 359 552 × 2 = 1 + 0.500 000 000 001 496 636 719 104;
  • 36) 0.500 000 000 001 496 636 719 104 × 2 = 1 + 0.000 000 000 002 993 273 438 208;
  • 37) 0.000 000 000 002 993 273 438 208 × 2 = 0 + 0.000 000 000 005 986 546 876 416;
  • 38) 0.000 000 000 005 986 546 876 416 × 2 = 0 + 0.000 000 000 011 973 093 752 832;
  • 39) 0.000 000 000 011 973 093 752 832 × 2 = 0 + 0.000 000 000 023 946 187 505 664;
  • 40) 0.000 000 000 023 946 187 505 664 × 2 = 0 + 0.000 000 000 047 892 375 011 328;
  • 41) 0.000 000 000 047 892 375 011 328 × 2 = 0 + 0.000 000 000 095 784 750 022 656;
  • 42) 0.000 000 000 095 784 750 022 656 × 2 = 0 + 0.000 000 000 191 569 500 045 312;
  • 43) 0.000 000 000 191 569 500 045 312 × 2 = 0 + 0.000 000 000 383 139 000 090 624;
  • 44) 0.000 000 000 383 139 000 090 624 × 2 = 0 + 0.000 000 000 766 278 000 181 248;
  • 45) 0.000 000 000 766 278 000 181 248 × 2 = 0 + 0.000 000 001 532 556 000 362 496;
  • 46) 0.000 000 001 532 556 000 362 496 × 2 = 0 + 0.000 000 003 065 112 000 724 992;
  • 47) 0.000 000 003 065 112 000 724 992 × 2 = 0 + 0.000 000 006 130 224 001 449 984;
  • 48) 0.000 000 006 130 224 001 449 984 × 2 = 0 + 0.000 000 012 260 448 002 899 968;
  • 49) 0.000 000 012 260 448 002 899 968 × 2 = 0 + 0.000 000 024 520 896 005 799 936;
  • 50) 0.000 000 024 520 896 005 799 936 × 2 = 0 + 0.000 000 049 041 792 011 599 872;
  • 51) 0.000 000 049 041 792 011 599 872 × 2 = 0 + 0.000 000 098 083 584 023 199 744;
  • 52) 0.000 000 098 083 584 023 199 744 × 2 = 0 + 0.000 000 196 167 168 046 399 488;
  • 53) 0.000 000 196 167 168 046 399 488 × 2 = 0 + 0.000 000 392 334 336 092 798 976;
  • 54) 0.000 000 392 334 336 092 798 976 × 2 = 0 + 0.000 000 784 668 672 185 597 952;

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:


0.000 000 000 742 147 676 646 753(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 753(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:


0.000 000 000 742 147 676 646 753(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 -0.000 000 000 742 147 676 646 753 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 646 714 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