-0.000 000 000 742 145 93 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 145 93(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
-0.000 000 000 742 145 93(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:

|-0.000 000 000 742 145 93| = 0.000 000 000 742 145 93


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: 0.000 000 000 742 145 93.

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) 0.000 000 000 742 145 93 × 2 = 0 + 0.000 000 001 484 291 86;
  • 2) 0.000 000 001 484 291 86 × 2 = 0 + 0.000 000 002 968 583 72;
  • 3) 0.000 000 002 968 583 72 × 2 = 0 + 0.000 000 005 937 167 44;
  • 4) 0.000 000 005 937 167 44 × 2 = 0 + 0.000 000 011 874 334 88;
  • 5) 0.000 000 011 874 334 88 × 2 = 0 + 0.000 000 023 748 669 76;
  • 6) 0.000 000 023 748 669 76 × 2 = 0 + 0.000 000 047 497 339 52;
  • 7) 0.000 000 047 497 339 52 × 2 = 0 + 0.000 000 094 994 679 04;
  • 8) 0.000 000 094 994 679 04 × 2 = 0 + 0.000 000 189 989 358 08;
  • 9) 0.000 000 189 989 358 08 × 2 = 0 + 0.000 000 379 978 716 16;
  • 10) 0.000 000 379 978 716 16 × 2 = 0 + 0.000 000 759 957 432 32;
  • 11) 0.000 000 759 957 432 32 × 2 = 0 + 0.000 001 519 914 864 64;
  • 12) 0.000 001 519 914 864 64 × 2 = 0 + 0.000 003 039 829 729 28;
  • 13) 0.000 003 039 829 729 28 × 2 = 0 + 0.000 006 079 659 458 56;
  • 14) 0.000 006 079 659 458 56 × 2 = 0 + 0.000 012 159 318 917 12;
  • 15) 0.000 012 159 318 917 12 × 2 = 0 + 0.000 024 318 637 834 24;
  • 16) 0.000 024 318 637 834 24 × 2 = 0 + 0.000 048 637 275 668 48;
  • 17) 0.000 048 637 275 668 48 × 2 = 0 + 0.000 097 274 551 336 96;
  • 18) 0.000 097 274 551 336 96 × 2 = 0 + 0.000 194 549 102 673 92;
  • 19) 0.000 194 549 102 673 92 × 2 = 0 + 0.000 389 098 205 347 84;
  • 20) 0.000 389 098 205 347 84 × 2 = 0 + 0.000 778 196 410 695 68;
  • 21) 0.000 778 196 410 695 68 × 2 = 0 + 0.001 556 392 821 391 36;
  • 22) 0.001 556 392 821 391 36 × 2 = 0 + 0.003 112 785 642 782 72;
  • 23) 0.003 112 785 642 782 72 × 2 = 0 + 0.006 225 571 285 565 44;
  • 24) 0.006 225 571 285 565 44 × 2 = 0 + 0.012 451 142 571 130 88;
  • 25) 0.012 451 142 571 130 88 × 2 = 0 + 0.024 902 285 142 261 76;
  • 26) 0.024 902 285 142 261 76 × 2 = 0 + 0.049 804 570 284 523 52;
  • 27) 0.049 804 570 284 523 52 × 2 = 0 + 0.099 609 140 569 047 04;
  • 28) 0.099 609 140 569 047 04 × 2 = 0 + 0.199 218 281 138 094 08;
  • 29) 0.199 218 281 138 094 08 × 2 = 0 + 0.398 436 562 276 188 16;
  • 30) 0.398 436 562 276 188 16 × 2 = 0 + 0.796 873 124 552 376 32;
  • 31) 0.796 873 124 552 376 32 × 2 = 1 + 0.593 746 249 104 752 64;
  • 32) 0.593 746 249 104 752 64 × 2 = 1 + 0.187 492 498 209 505 28;
  • 33) 0.187 492 498 209 505 28 × 2 = 0 + 0.374 984 996 419 010 56;
  • 34) 0.374 984 996 419 010 56 × 2 = 0 + 0.749 969 992 838 021 12;
  • 35) 0.749 969 992 838 021 12 × 2 = 1 + 0.499 939 985 676 042 24;
  • 36) 0.499 939 985 676 042 24 × 2 = 0 + 0.999 879 971 352 084 48;
  • 37) 0.999 879 971 352 084 48 × 2 = 1 + 0.999 759 942 704 168 96;
  • 38) 0.999 759 942 704 168 96 × 2 = 1 + 0.999 519 885 408 337 92;
  • 39) 0.999 519 885 408 337 92 × 2 = 1 + 0.999 039 770 816 675 84;
  • 40) 0.999 039 770 816 675 84 × 2 = 1 + 0.998 079 541 633 351 68;
  • 41) 0.998 079 541 633 351 68 × 2 = 1 + 0.996 159 083 266 703 36;
  • 42) 0.996 159 083 266 703 36 × 2 = 1 + 0.992 318 166 533 406 72;
  • 43) 0.992 318 166 533 406 72 × 2 = 1 + 0.984 636 333 066 813 44;
  • 44) 0.984 636 333 066 813 44 × 2 = 1 + 0.969 272 666 133 626 88;
  • 45) 0.969 272 666 133 626 88 × 2 = 1 + 0.938 545 332 267 253 76;
  • 46) 0.938 545 332 267 253 76 × 2 = 1 + 0.877 090 664 534 507 52;
  • 47) 0.877 090 664 534 507 52 × 2 = 1 + 0.754 181 329 069 015 04;
  • 48) 0.754 181 329 069 015 04 × 2 = 1 + 0.508 362 658 138 030 08;
  • 49) 0.508 362 658 138 030 08 × 2 = 1 + 0.016 725 316 276 060 16;
  • 50) 0.016 725 316 276 060 16 × 2 = 0 + 0.033 450 632 552 120 32;
  • 51) 0.033 450 632 552 120 32 × 2 = 0 + 0.066 901 265 104 240 64;
  • 52) 0.066 901 265 104 240 64 × 2 = 0 + 0.133 802 530 208 481 28;
  • 53) 0.133 802 530 208 481 28 × 2 = 0 + 0.267 605 060 416 962 56;
  • 54) 0.267 605 060 416 962 56 × 2 = 0 + 0.535 210 120 833 925 12;

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:


0.000 000 000 742 145 93(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1000 00(2)

6. Positive number before normalization:

0.000 000 000 742 145 93(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1000 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:


0.000 000 000 742 145 93(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1000 00(2) × 20 =

1.1001 0111 1111 1111 1100 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 0111 1111 1111 1100 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 1011 1111 1111 1110 0000 =

100 1011 1111 1111 1110 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 1011 1111 1111 1110 0000


Decimal number -0.000 000 000 742 145 93 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1110 0000

Share

» Convert decimal -0.000 000 000 742 145 64 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: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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